[{"data":1,"prerenderedAt":1110},["ShallowReactive",2],{"layer:constructing-angles:deepen":3},{"layer":4,"contentHash":1083,"dependencyHashes":1084,"approval":1103,"releaseId":1109},{"schemaVersion":5,"conceptId":6,"locale":7,"depth":8,"revision":5,"title":9,"subtitle":10,"summary":11,"objectives":12,"estimatedMinutes":18,"plate":19,"blocks":40,"sourceIds":1078,"reviewStatus":1079,"authoring":1080},1,"constructing-angles","en","deepen","Why the constructions work","Proofs behind the recipes, edge cases, accuracy and the problems the Greeks could not solve","Find out why each compass construction is exact: equilateral triangles for 60°, congruent triangles for bisectors, equidistant points for perpendiculars. Then test edge cases, measure reflex angles, analyse errors and meet the impossible trisection problem.",[13,14,15,16,17],"Explain with equal radii and congruent triangles why the 60°, bisector, perpendicular bisector and copy-angle constructions are exact.","Use the average rule and straight lines to plan the shortest route to 15°, 75°, 105°, 135°, 150° and 165°.","Spot and explain edge cases, such as arcs that are too small to meet.","Measure reflex angles two ways and estimate how drawing errors change a measured angle.","Describe why compass and straightedge alone were chosen, and what it means that trisection is impossible.",45,{"title":20,"rows":21},"Lesson plate",[22,25,28,31,34,37],{"label":23,"value":24},"Depth","Go deeper",{"label":26,"value":27},"Reading time","≈ 45 minutes",{"label":29,"value":30},"Prior knowledge","The basic constructions (Understand)",{"label":32,"value":33},"Chapters","11",{"label":35,"value":36},"Labs","Reflex protractor, 120° and copy-angle animations, 2 games",{"label":38,"value":39},"Key tools","Equal radii, SSS, equidistant points",[41,45,51,57,60,65,70,99,104,109,112,126,131,134,143,147,174,179,182,187,197,200,212,233,237,242,245,248,257,262,276,281,302,320,323,327,332,339,342,364,368,372,377,380,432,442,451,464,485,499,504,507,512,516,537,541,546,549,557,575,578,602,615,620,625,628,641,645,685,689,694,731,801,805,874,1030,1046,1052,1057,1062],{"id":42,"type":43,"markdown":44},"intro","prose","In earlier layers you learned *how* to construct 60°, 90°, 30°, 45°, bisectors and perpendiculars. You followed the recipes and checked them with a protractor. This layer asks the question a mathematician asks: **why do the recipes work, every single time, for any size of compass opening?**\n\nA protractor reading can only ever tell you that an angle is *about* 60°, within a degree or so. A construction, backed by a reason, tells you it is **exactly** 60°, not 59.9° and not 60.1°. That exactness is the whole point of the compass-and-straightedge game, and it is why these constructions were studied for more than two thousand years.\n\nYou will meet three powerful tools of reasoning along the way: **equal radii** (every point on an arc is the same distance from its centre), **congruent triangles** (two triangles with the same three sides fit exactly on top of each other) and **equidistant points** (points the same distance from two places). With those three ideas, every construction in this topic can be explained.",{"id":46,"type":47,"variant":48,"title":49,"markdown":50},"how-to","callout","observation","How to read this layer","Each chapter states a construction, then gives the reason it works, then tests the reason on an edge case: what happens if the compass is too small, the angle is reflex, or your hand slips. Keep a compass, ruler, pencil and protractor beside you and redraw each figure. The letters match the animations and the earlier layers: the 60° construction uses O, A, P, Q; the angle bisector uses O, A, B, P, Q and T; the copy uses O′, A′, P′, Q′.",{"id":52,"type":53,"title":54,"eyebrow":55,"navLabel":56},"ch1","chapter","The rules of the game: why only compass and straightedge?","Chapter 01","1 Rules of the game",{"id":58,"type":43,"markdown":59},"rules-prose","Ancient Greek geometers, whose work was collected by **Euclid in the Elements (about 300 BCE)**, allowed themselves just two tools:\n\n- a **straightedge**: it draws a straight line through two points you already have. It has **no markings**, so you cannot measure with it.\n- a **compass**: it draws a circle with a chosen centre passing through a chosen point.\n\nEverything else must be *built* from those two moves. Why so strict? Because the Greeks wanted geometry to rest on the fewest possible assumptions. A line and a circle are the simplest perfect shapes. If a result can be reached using only them, then it is true because of logic, not because somebody's ruler was printed accurately. A construction was really a **proof in pictures**: each step comes with a reason, and the final figure is guaranteed.\n\nIn school we use a ruler with centimetre marks, but in a construction we use it only as a straightedge. The marks are allowed for *checking* afterwards, never for building.",{"id":61,"type":47,"variant":62,"title":63,"markdown":64},"collapsing","nuance","Euclid's \"collapsing\" compass","Euclid's compass was imagined to snap shut the moment you lifted it off the page. You could draw a circle, but you could not carry a distance from one place to another. Surprisingly, this does not matter. Euclid's second proposition (Book I, Prop. 2) shows how to copy a length using only a collapsing compass and a straightedge, with a clever equilateral-triangle trick. So a modern compass that holds its width can do nothing a collapsing compass cannot; it is just quicker. When you copy a segment with a fixed compass opening, you are using a shortcut Euclid proved was allowed.",{"id":66,"type":47,"variant":67,"title":68,"markdown":69},"sulba","example","India's rope geometry","Long before Euclid, the **Sulba Sutras** (the oldest parts are often dated to around 800–500 BCE) gave rules for building Vedic fire altars of exact shapes and areas. Their tools were a **rope (sulba) and pegs**: a peg with a rope tied to it is a compass, and a rope stretched tight between two pegs is a straightedge. The texts describe how to make a right angle by marking a rope in lengths like 3, 4 and 5 units and pegging it into a triangle, and how to find the east-west line and its perpendicular. The same two tools, the same idea: build exact shapes from straight lines and circles.",{"id":71,"type":72,"tone":73,"items":74},"tool-rules","spec","blue",[75,79,83,87,91,95],{"label":76,"big":77,"value":78},"Straightedge may","join","Draw the line through two points you already have, and extend a segment as far as you like.",{"label":80,"big":81,"value":82},"Straightedge may not","measure","No reading of centimetres while constructing; marks are for checking only.",{"label":84,"big":85,"value":86},"Compass may","circle","Draw a circle or arc with a known centre through a known point, or with a width you have copied.",{"label":88,"big":89,"value":90},"Compass may not","guess","Set its width to a number like 3.7 cm \"by eye\" and call that exact.",{"label":92,"big":93,"value":94},"New points come from","crossings","Where lines and circles cross each other. Nothing else creates a point.",{"label":96,"big":97,"value":98},"A result is exact if","reasoned","Every step has a reason, so the conclusion holds for any size of drawing.",{"id":100,"type":47,"variant":101,"title":102,"markdown":103},"mis-radius","misconception","\"The construction only works with the radius in the book\"","Textbooks often say \"take a radius of 4 cm\", but the number is only a suggestion to keep the drawing a sensible size. The reasoning never uses the value 4. That is exactly what makes a construction powerful: the 60° recipe gives 60° with a radius of 2 cm, 5 cm or 2 metres chalked on a playground. What *does* matter is that you keep the radius the same where the recipe says \"same radius\".",{"id":105,"type":53,"title":106,"eyebrow":107,"navLabel":108},"ch2","Why the 60° construction is exactly 60°","Chapter 02","2 Why 60° works",{"id":110,"type":43,"markdown":111},"sixty-recipe","Recall the recipe. Draw a ray OA. With centre O and any radius, draw an arc cutting OA at P. **Keeping the same radius**, put the compass point on P and draw an arc cutting the first arc at Q. Join OQ. Then ∠QOA = 60°.\n\nNow the reason. Look at the three points O, P and Q and the three distances between them:\n\n- **OP** is a radius of the first arc (centre O).\n- **OQ** is also a radius of the first arc, because Q lies on that arc.\n- **PQ** is a radius of the second arc (centre P), and that arc had the **same radius**.\n\nSo OP = OQ = PQ. Triangle OPQ has three equal sides: it is **equilateral**.",{"id":113,"type":114,"title":115,"problem":116,"steps":117,"help":124},"we-equilateral","worked_example","Why every angle of an equilateral triangle is 60°","Triangle OPQ has OP = OQ = PQ. Show that ∠QOP = 60°.",[118,119,120,121,122,123],"In any triangle, the angles opposite equal sides are equal. (Fold an isosceles triangle along its line of symmetry and the two base angles land on each other.)","OP = OQ, so the angles opposite them are equal: ∠OQP = ∠OPQ.","OQ = PQ, so the angles opposite them are equal: ∠OPQ = ∠QOP.","Put these together: all three angles are equal. Call each one x.","The angles of a triangle add up to 180°, so x + x + x = 180°, which means 3x = 180°.","x = 180° ÷ 3 = **60°**. So ∠QOP = ∠QOA = 60°, exactly, whatever radius you chose.",{"simplerExplanation":125},"The recipe secretly draws a triangle with three equal sides. Three equal sides force three equal angles, and three equal angles that add up to 180° must each be 60°.",{"id":127,"type":47,"variant":128,"title":129,"markdown":130},"aha-hidden","aha","The triangle you never drew","You only drew two arcs and one line, yet the proof talks about triangle OPQ. The side PQ is invisible on your page! Constructions often work because of shapes hidden in the figure. When you want to understand a construction, join up the points you marked and look for equal lengths.",{"id":132,"type":43,"markdown":133},"hexagon-prose","Here is a beautiful consequence. Keep the compass at the same width and walk it round a circle: centre on P, mark Q; centre on Q, mark R; and so on. Each step cuts off a chord equal to the radius, so each step makes an equilateral triangle with the centre, and each step turns through **60°** at the centre.\n\nAfter six steps the total turn is 6 × 60° = **360°**, a complete turn, so the sixth mark lands **exactly** back on the starting point. That is why a compass flower (the six-petal rangoli design) closes perfectly, and why joining the six marks gives a **regular hexagon** whose side equals the radius.",{"id":135,"type":114,"title":136,"problem":137,"steps":138},"we-hexagon","A hexagon from one compass setting","A compass is set to 4 cm and walked round a circle of radius 4 cm. Joining the marks gives a hexagon. How long is each side, what is its perimeter, and what angle does each side make at the centre?",[139,140,141,142],"Each step cuts the circle at a point 4 cm from the previous mark, so each side of the hexagon is **4 cm**.","Perimeter = 6 × 4 cm = **24 cm**.","Each side, together with two radii, forms an equilateral triangle, so each side subtends **60°** at the centre.","Check: 6 × 60° = 360°, one full turn, so the figure closes. ✓",{"id":144,"type":47,"variant":62,"title":145,"markdown":146},"nuance-drift","Exact in theory, a hair off in practice","On paper the sixth step sometimes misses the starting point by a millimetre. The geometry is not wrong: your pencil has width, the compass may slip a little, and every small error is carried forward to the next step, so errors **add up** over six steps. This is a useful lesson in the difference between a *proof* (exact) and a *drawing* (a physical model of the proof). A sharp pencil and a stiff compass keep the gap tiny.",{"id":148,"type":149,"itemId":150,"prompt":151,"check":152,"hints":168,"feedback":171},"pr-hexagon","practice","constructing-angles.dee-hexagon-steps","You walk a compass round a circle using the radius as the step, but you stop after **four** steps. Joining the centre O to the first and last marks, what angle do they make at O?",{"kind":153,"options":154,"correct":167},"choice",[155,158,161,164],{"id":156,"label":157},"a","180°",{"id":159,"label":160},"b","240°, measured the long way round (120° the short way)",{"id":162,"label":163},"c","200°",{"id":165,"label":166},"d","It depends on the radius",[159],[169,170],"Each step turns 60° at the centre.","4 × 60° = ?",{"correct":172,"incorrect":173},"Right: 4 × 60° = 240°, which is a reflex angle. The other way round the gap is 360° − 240° = 120°.","Each step adds 60° at the centre, whatever the radius. Four steps make 4 × 60° = 240°; the remaining gap is 360° − 240° = 120°.",{"id":175,"type":53,"title":176,"eyebrow":177,"navLabel":178},"ch3","Why the angle bisector splits an angle exactly in half","Chapter 03","3 Why bisecting works",{"id":180,"type":43,"markdown":181},"bis-recipe","The recipe for bisecting ∠AOB: with centre O draw an arc cutting the arms at P and Q. With centres P and Q and **equal radii**, draw two arcs that cross at T inside the angle. Join OT. Then OT bisects ∠AOB.\n\nTo see why, join P to T and Q to T. Now there are two triangles, **OPT** and **OQT**. Compare their sides:\n\n- **OP = OQ**: both are radii of the first arc, centre O.\n- **PT = QT**: both are radii of the two equal arcs.\n- **OT = OT**: the two triangles share this side.\n\nThree pairs of equal sides. That is the **SSS rule**: if two triangles have all three sides equal, they are **congruent**, meaning one fits exactly on top of the other (possibly after flipping it over). Congruent triangles have equal matching angles, so ∠POT = ∠QOT. OT cuts ∠AOB into two equal halves.",{"id":183,"type":47,"variant":184,"title":185,"markdown":186},"def-congruent","definition","Congruent and the SSS rule","Two shapes are **congruent** if one can be placed exactly on the other by sliding, turning or flipping: same size, same shape. For triangles, the **SSS (side-side-side) rule** says: if the three sides of one triangle equal the three sides of another, the triangles are congruent. Why should that be true? Because three lengths fix a triangle completely: once you know them, there is only one triangle you can build (you will build triangles from three sides in Extend). So the angles are fixed too.",{"id":188,"type":114,"title":189,"problem":190,"steps":191},"we-bis-reason","The bisector reason, step by step","∠AOB = 72°. Using the recipe, what are ∠AOT and ∠TOB, and which fact gives the answer without measuring?",[192,193,194,195,196],"OP = OQ (same arc, centre O). PT = QT (equal arcs from P and Q). OT is shared.","So triangles OPT and OQT are congruent by SSS.","Matching angles of congruent triangles are equal: ∠POT = ∠QOT.","These two angles together make ∠AOB = 72°, so each one is 72° ÷ 2 = **36°**.","No protractor needed: the answer follows from the equal lengths alone.",{"id":198,"type":43,"markdown":199},"bis-edge","**Edge case 1: the arcs do not meet.** The second pair of arcs, centred at P and Q, only cross if their radius is **more than half of PQ**. If the radius is exactly half of PQ, the arcs just touch at the midpoint of PQ (that still works, but it is hard to see). If it is less than half, the arcs never meet and there is no T at all.\n\nHow long is PQ? If the first arc has radius r and the angle is θ, then triangle OPQ is isosceles and PQ grows with the angle. For a 60° angle, triangle OPQ is equilateral, so PQ = r. The safe rule in textbooks, \"take a radius more than half of PQ\", comes straight from this.\n\n**Edge case 2: which crossing point?** Two circles that cross meet in **two** points, one on each side of the line PQ. Both lie on the bisector line, because each is equally far from P and from Q. So either one works. If you reuse the very first radius, one crossing point is O itself (O is already r from both P and Q) and the useful one is the other: that is fine, just do not join O to O!",{"id":201,"type":114,"title":202,"problem":203,"steps":204,"help":210},"we-bis-radius","How wide must the second compass opening be?","∠AOB = 50° and the first arc has radius 5 cm. What is the smallest radius for the arcs from P and Q that still lets them cross?",[205,206,207,208,209],"Triangle OPQ has OP = OQ = 5 cm and apex angle 50°. Split it down the middle: each half is a right-angled triangle with a 25° angle at O.","Half of PQ = 5 × sin 25°. (If you have not met sine yet, just draw it accurately and measure: you get the same.)","sin 25° ≈ 0.4226, so half of PQ ≈ 5 × 0.4226 ≈ **2.11 cm**, and PQ ≈ 4.23 cm.","The second radius must be **more than about 2.11 cm**. A comfortable choice is 3 or 4 cm.","For a wider angle, PQ is longer, so you need a wider second opening.",{"simplerExplanation":211},"The two arcs are like two people walking towards each other from P and from Q. If each can only walk less than halfway, they never meet.",{"id":213,"type":149,"itemId":214,"prompt":215,"check":216,"hints":227,"feedback":230},"pr-bis-radius","constructing-angles.dee-bisector-radius","You bisect a **60°** angle. Your first arc, centre O, has radius 6 cm, cutting the arms at P and Q. Which radius for the arcs from P and Q will **fail** to give a crossing point?",{"kind":153,"options":217,"correct":226},[218,220,222,224],{"id":156,"label":219},"2.5 cm",{"id":159,"label":221},"3.5 cm",{"id":162,"label":223},"5 cm",{"id":165,"label":225},"6 cm",[156],[228,229],"For a 60° angle, triangle OPQ is equilateral. How long is PQ?","The arcs need a radius more than half of PQ.",{"correct":231,"incorrect":232},"Right. Triangle OPQ is equilateral, so PQ = 6 cm and the arcs need a radius more than 3 cm. 2.5 cm is too short.","With a 60° angle, OP = OQ = 6 cm and the apex is 60°, so OPQ is equilateral and PQ = 6 cm. The arcs need radius more than 3 cm, so only 2.5 cm fails. (6 cm works: one crossing is O itself, the other is on the bisector.)",{"id":234,"type":47,"variant":101,"title":235,"markdown":236},"mis-bisect-arms","\"Bisecting means halving the arms\"","Some learners mark the midpoint of each arm and join the midpoints to the vertex. That does not bisect the angle unless the two arms happen to be the same length, and even then only by luck. The arms of an angle are rays: they go on for ever, so they do not have lengths or midpoints. The bisector is about the **opening**, not the arms, which is why the recipe starts by cutting off **equal** lengths OP and OQ with one arc.",{"id":238,"type":53,"title":239,"eyebrow":240,"navLabel":241},"ch4","Why the perpendicular bisector is perpendicular and bisects","Chapter 04","4 Perpendicular bisector",{"id":243,"type":43,"markdown":244},"pb-recipe","To draw the perpendicular bisector of segment AB: open the compass to **more than half of AB**. With centre A draw arcs above and below AB; with the same radius and centre B draw arcs that cross them at P (above) and Q (below). Join PQ, meeting AB at M. Then PQ is perpendicular to AB and M is the midpoint of AB.\n\nThe reason: AP = BP = AQ = BQ, all equal to the one radius you used. A four-sided shape with four equal sides is a **rhombus**, so APBQ is a rhombus. The two diagonals of a rhombus always **cut each other in half at right angles**. The diagonals here are AB and PQ, so PQ meets AB at its midpoint, at 90°.",{"id":246,"type":43,"markdown":247},"pb-locus","There is an even deeper way to see it. P is the same distance from A as from B; we say P is **equidistant** from A and B. So is Q. So is M. In fact:\n\n> **Every point on the perpendicular bisector of AB is equidistant from A and B, and every point equidistant from A and B lies on the perpendicular bisector.**\n\nThe set of all points that obey a rule is called a **locus**. The perpendicular bisector is the locus of points equidistant from A and B. The construction simply finds two such points (P and Q) and draws the line through them. Since two points fix a straight line, the whole bisector is found.\n\nThat is why you may use a **different** radius above and below AB: P could come from 4 cm arcs and Q from 6 cm arcs. APBQ is then a **kite**, not a rhombus, but P and Q are still each equidistant from A and B, so PQ is still the perpendicular bisector.",{"id":249,"type":114,"title":250,"problem":251,"steps":252},"we-pb-kite","Proving it with the kite version","P is 4 cm from both A and B, and Q (on the other side of AB) is 6 cm from both. Explain why PQ ⟂ AB and AM = MB.",[253,254,255,256],"Triangles APQ and BPQ: AP = BP = 4 cm, AQ = BQ = 6 cm, and PQ is shared. So they are congruent (SSS), which gives ∠APM = ∠BPM.","Now triangles APM and BPM: AP = BP, ∠APM = ∠BPM, and PM is shared. Two sides and the angle between them match (the **SAS rule**), so these triangles are congruent too.","Therefore AM = BM: M is the midpoint.","Also ∠AMP = ∠BMP. These two angles sit side by side on the straight line AB, so they add to 180°. Equal and adding to 180° means each is **90°**.",{"id":258,"type":47,"variant":259,"title":260,"markdown":261},"try-town","try_it","Where should the well go?","Two villages, A and B, want to share a new water pump placed the same distance from both. Mark two dots 8 cm apart and construct the perpendicular bisector. Every point on that line is fair to both villages. Add a third village C, not on the line AB. Construct the perpendicular bisector of BC as well. Where the two bisectors cross is **one** point equidistant from all three villages: the centre of the circle through A, B and C. Check it by drawing that circle.",{"id":263,"type":264,"prompt":265,"options":266,"explanation":275},"pred-pb","prediction","In the perpendicular bisector recipe, you accidentally open the compass to **less than half of AB**. What happens?",[267,269,271,273],{"id":156,"label":268},"The line comes out slightly slanted",{"id":159,"label":270},"The arcs from A and B never cross, so there are no points P and Q",{"id":162,"label":272},"It still works, only the line is shorter",{"id":165,"label":274},"M lands closer to A than to B","**The arcs never meet.** A point equidistant from A and B at distance r can only exist if 2r is at least AB; otherwise the two circles are too small to reach each other. There is no slanted wrong answer: the method either works exactly or gives nothing at all. That all-or-nothing behaviour is typical of exact constructions.",{"id":277,"type":53,"title":278,"eyebrow":279,"navLabel":280},"ch5","Perpendiculars at a point and from a point","Chapter 05","5 Perpendiculars",{"id":282,"type":283,"title":284,"items":285},"steps-perp-at","steps","Perpendicular to a line at a point P on it",[286,290,294,298],{"title":287,"tag":288,"text":289},"Equal cuts","step 1","With centre P and any radius, draw an arc cutting the line at X and Y. Now PX = PY, so P is the midpoint of XY.",{"title":291,"tag":292,"text":293},"Wider arcs","step 2","Open the compass wider (more than PX). With centres X and Y, draw arcs that cross at Z above the line.",{"title":295,"tag":296,"text":297},"Join","step 3","Join PZ. It is perpendicular to the line at P.",{"title":299,"tag":300,"text":301},"Why","reason","Z is equidistant from X and Y, so Z is on the perpendicular bisector of XY. That bisector passes through P, the midpoint. So PZ is it.",{"id":303,"type":283,"title":304,"items":305},"steps-perp-from","Perpendicular from a point Z not on the line",[306,309,312,314,316],{"title":307,"tag":288,"text":308},"Cut the line","With centre Z, draw an arc cutting the line at two points X and Y. Now ZX = ZY.",{"title":310,"tag":292,"text":311},"Arcs below","With centres X and Y and equal radii, draw arcs crossing at W on the other side of the line.",{"title":295,"tag":296,"text":313},"Join ZW. It meets the line at a right angle, at a point N.",{"title":299,"tag":300,"text":315},"Z and W are both equidistant from X and Y, so both lie on the perpendicular bisector of XY. Two points fix the line, so ZW is that bisector.",{"title":317,"tag":318,"text":319},"Bonus","shortest path","ZN is the shortest distance from Z to the line: any other path to the line is the long side of a right-angled triangle.",{"id":321,"type":43,"markdown":322},"perp-prose","Notice that both recipes are really the **perpendicular bisector construction in disguise**. The first step manufactures a segment XY whose perpendicular bisector is the line you want; the second step finds that bisector. Once you see this, you do not need to memorise three separate recipes, only one idea: *find two points equidistant from X and Y*.\n\nThe second recipe also answers a practical question: how far is a point from a road? The **distance from a point to a line** always means the perpendicular distance, ZN, because it is the shortest.",{"id":324,"type":47,"variant":101,"title":325,"markdown":326},"mis-shortest","\"Any line from the point to the road gives the distance\"","A slanted line from Z to the road is always longer than the perpendicular ZN, because in a right-angled triangle the side opposite the right angle (the hypotenuse) is the longest side. That is why a surveyor, a map-maker or a traffic engineer measures distance to a road along the perpendicular.",{"id":328,"type":53,"title":329,"eyebrow":330,"navLabel":331},"ch6","Why 90° comes from 60° and 120°","Chapter 06","6 Why 90° works",{"id":333,"type":334,"component":335,"componentVersion":5,"config":336,"textAlternative":338},"anim-120","animation","compass-construction",{"construction":337},"angle-120","This animation constructs a 120° angle at O on ray OA, and shows why it is exact.\n\n1. A ray OA is drawn.\n2. With centre O and a comfortable radius, an arc is drawn from the ray upward and round to the left, cutting OA at P.\n3. Keeping the same radius, the compass point moves to P and a small arc cuts the big arc at Q. Triangle OPQ is equilateral (OP = OQ = PQ = the radius), so ∠POQ = 60°.\n4. Keeping the same radius again, the compass point moves to Q and cuts the big arc at R. Triangle OQR is also equilateral, so ∠QOR = 60°.\n5. The ray OR is drawn. ∠AOR = ∠AOQ + ∠QOR = 60° + 60° = **120°**.\n\nWhy it is exact: two equilateral triangles sit side by side at O, each contributing exactly 60°. The radius never changes, so the reasoning holds for any size of drawing. If you continued one more step, you would reach 180°, landing exactly on the ray opposite OA, which is a good check on your accuracy.",{"id":340,"type":43,"markdown":341},"ninety-reason","With the 60° mark Q and the 120° mark R on the same arc, the 90° recipe says: with centres Q and R and equal radii, draw arcs crossing at S. Join OS. Why is ∠AOS exactly 90°?\n\nLook at what you have. O is equidistant from Q and R (both are on the first arc). S is equidistant from Q and R (equal arcs). So O and S both lie on the perpendicular bisector of QR, and **OS bisects ∠QOR**. Since ∠QOR = 120° − 60° = 60°, the bisector splits it into 30° + 30°. Therefore\n\n∠AOS = 60° + 30° = **90°**.\n\nYou can also see it as an average: the bisector of the gap between the 60° ray and the 120° ray points to (60° + 120°) ÷ 2 = 90°. This \"average rule\" is the key to every combination angle in the next chapter.",{"id":343,"type":344,"items":345},"f-average","formulas",[346,349,352,355,358,361],{"expression":347,"caption":348},"bisect(a, b) = (a + b) ÷ 2","Bisecting the gap between rays at a° and b° (from the same base ray) gives the ray at their average.",{"expression":350,"caption":351},"(60 + 120) ÷ 2 = 90","The usual 90° recipe.",{"expression":353,"caption":354},"(0 + 180) ÷ 2 = 90","Bisecting a straight angle: the perpendicular-at-a-point recipe.",{"expression":356,"caption":357},"(0 + 60) ÷ 2 = 30","Bisecting the 60° angle itself.",{"expression":359,"caption":360},"(60 + 90) ÷ 2 = 75","The 75° recipe.",{"expression":362,"caption":363},"180 − a","Measuring from the other end of a straight line turns a into its supplement.",{"id":365,"type":47,"variant":62,"title":366,"markdown":367},"nuance-two-ways","Two routes to 90°, same destination","Route 1 bisects the 60° gap between the 60° and 120° marks. Route 2 extends OA backwards to make a straight angle (180°) and bisects that, which is exactly the perpendicular-at-a-point recipe. Both give exact 90°. Route 2 uses fewer arcs, but Route 1 keeps everything on one side of the line and gives you 60° and 120° as a bonus, which is handy when you next want 75° or 105°.",{"id":369,"type":47,"variant":128,"title":370,"markdown":371},"aha-linear","Bisectors of a linear pair are always perpendicular","Draw any line through O and a ray making a linear pair, say x° and (180 − x)°. Bisect both angles. The two bisectors make x\u002F2 + (180 − x)\u002F2 = 180\u002F2 = **90°**, whatever x is. Try it with x = 40°: 20° + 70° = 90°. With x = 115°: 57.5° + 32.5° = 90°. The x cancels out, so this is always true.",{"id":373,"type":53,"title":374,"eyebrow":375,"navLabel":376},"ch7","Recipes: building 15°, 75°, 105°, 135°, 150°, 165° and more","Chapter 07","7 Angle recipes",{"id":378,"type":43,"markdown":379},"recipe-intro","With exact 60° steps, exact bisection and the straight line (180°), you can reach a large family of angles. Every recipe is a combination of three moves:\n\n1. **Step** 60° along an arc (to 60°, 120°, 180°).\n2. **Bisect** the gap between two rays you already have (the average rule).\n3. **Use the straight line**: an angle x on one side of a point gives 180° − x on the other.\n\nA good constructor looks for the **shortest route**, because every extra arc is another chance for a small error.",{"id":381,"type":382,"caption":383,"columns":384,"rows":388},"t-recipes","table","Recipes for constructible angles (all on base ray OA, vertex O)",[385,386,387],"Angle","Route","Why it is exact",[389,393,397,401,405,408,412,416,420,424,428],[390,391,392],"15°","Construct 30° (bisect 60°), then bisect again","60 ÷ 2 ÷ 2 = 15",[394,395,396],"22.5°","Construct 90°, bisect to 45°, bisect again","90 ÷ 2 ÷ 2 = 22.5",[398,399,400],"30°","Bisect the 60° angle","60 ÷ 2 = 30",[402,403,404],"45°","Construct 90°, bisect it","90 ÷ 2 = 45",[406,407,359],"75°","Bisect between the 60° and 90° rays",[409,410,411],"105°","Bisect between the 90° and 120° rays","(90 + 120) ÷ 2 = 105",[413,414,415],"135°","Bisect between the 90° ray and the extended line (180°)","(90 + 180) ÷ 2 = 135, also 180 − 45",[417,418,419],"150°","Bisect between the 120° ray and the extended line","(120 + 180) ÷ 2 = 150, also 180 − 30",[421,422,423],"165°","Bisect between the 150° ray and the extended line","(150 + 180) ÷ 2 = 165, also 180 − 15",[425,426,427],"7.5°","Bisect a 15° angle","15 ÷ 2 = 7.5",[429,430,431],"37.5°","Bisect between the 30° and 45° rays","(30 + 45) ÷ 2 = 37.5",{"id":433,"type":114,"title":434,"problem":435,"steps":436},"we-75","Constructing 75° and proving it","Construct ∠AOX = 75° and explain why it is exact.",[437,438,439,440,441],"Draw ray OA. Arc centre O cuts OA at P. Same radius from P gives Q (60° mark); from Q gives R (120° mark).","Arcs from Q and R with equal radii cross at S. Ray OS is at 90° (bisector of the 60°–120° gap).","OS cuts the big arc at T. Now bisect ∠QOT: arcs from Q and T with equal radii cross at U.","Ray OX through U bisects the gap between 60° and 90°, so ∠AOX = (60° + 90°) ÷ 2 = **75°**.","Check with a protractor: it should read 75° within about a degree.",{"id":443,"type":114,"title":444,"problem":445,"steps":446},"we-135","Two proofs that a route gives 135°","Ray OA is extended backwards to A′ to make a straight line. OS is perpendicular (∠AOS = 90°). OX bisects ∠SOA′. Find ∠AOX two ways.",[447,448,449,450],"∠SOA′ = 180° − 90° = 90°, so its bisector makes ∠SOX = 45°.","Way 1: ∠AOX = ∠AOS + ∠SOX = 90° + 45° = **135°**.","Way 2 (average rule): OX is halfway between the 90° ray and the 180° ray, so ∠AOX = (90° + 180°) ÷ 2 = **135°**.","Way 3 (straight line): ∠XOA′ = 45°, so ∠AOX = 180° − 45° = **135°**. Three routes, one answer.",{"id":452,"type":264,"prompt":453,"options":454,"explanation":463},"pred-15s","Using only 60° steps, bisecting and the straight line, which of these can you **reach exactly**?",[455,457,459,461],{"id":156,"label":456},"Every multiple of 15° from 15° to 180°",{"id":159,"label":458},"Only 30°, 45°, 60°, 90° and 120°",{"id":162,"label":460},"Every whole number of degrees",{"id":165,"label":462},"Every multiple of 10°","**Every multiple of 15° from 15° to 180°**, twelve angles in all: 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165 and 180. The table above shows a route to each. Bisecting further gives halves such as 7.5°, 22.5° and 37.5°. But **20°, 40°, 50°, 70° and 80° never appear**, however long you keep bisecting and combining: these moves always produce multiples of 15° divided by a power of 2. Whether some other clever trick could reach 20° is a much harder question, answered in chapter 10.",{"id":465,"type":149,"itemId":466,"prompt":467,"check":468,"hints":479,"feedback":482},"pr-165","constructing-angles.dee-route-165","Which route gives exactly **165°** at O on ray OA?",{"kind":153,"options":469,"correct":478},[470,472,474,476],{"id":156,"label":471},"Bisect between the 150° ray and the extended line OA′",{"id":159,"label":473},"Bisect between the 120° ray and the 180° line, then add 15° by eye",{"id":162,"label":475},"Step 60° three times, then bisect",{"id":165,"label":477},"Bisect between the 135° and 180° rays",[156],[480,481],"Use the average rule (a + b) ÷ 2.","Which pair of rays averages to 165?",{"correct":483,"incorrect":484},"Right: (150° + 180°) ÷ 2 = 165°, or 180° − 15°.","Use (a + b) ÷ 2: (150 + 180) ÷ 2 = 165. Option d gives (135 + 180) ÷ 2 = 157.5°; option c gives 90°; option b adds an estimate, which is not a construction.",{"id":486,"type":149,"itemId":487,"prompt":488,"check":489,"hints":494,"feedback":496},"pr-average","constructing-angles.dee-average-rule","Rays OX and OY are constructed at 105° and 150° from OA. You bisect ∠XOY. How many degrees from OA is the bisector?",{"kind":490,"answer":491,"tolerance":492,"unit":493},"number",127.5,0,"°",[495],"The bisector sits at the average of the two rays.",{"correct":497,"incorrect":498},"Right: (105 + 150) ÷ 2 = 255 ÷ 2 = 127.5°.","Average the two directions: (105 + 150) ÷ 2 = 127.5°. The gap is 45°, and half of it is 22.5°, so 105 + 22.5 = 127.5°.",{"id":500,"type":53,"title":501,"eyebrow":502,"navLabel":503},"ch8","Copying an angle, and why it works","Chapter 08","8 Copying an angle",{"id":505,"type":43,"markdown":506},"copy-intro","Sometimes you are given an angle with **no** idea of its size: a carpenter's corner, a roof pitch, an angle drawn by a friend. Can you make an exact copy of it somewhere else, using no protractor? Yes. The trick is that an angle is completely fixed by an isosceles triangle cut from it: two equal arms and the distance between their ends.",{"id":508,"type":334,"component":335,"componentVersion":5,"config":509,"textAlternative":511},"anim-copy",{"construction":510},"copy-angle","This animation copies an angle ∠AOB onto a new ray O′A′, using only compass and straightedge.\n\n1. The given angle ∠AOB (50° in the animation) is shown, with vertex O. A separate ray O′A′ is drawn where the copy will go.\n2. With centre O and any radius, an arc cuts the arms of ∠AOB at P (on OA) and Q (on OB).\n3. Without changing the radius, the compass point goes to O′ and a long arc is drawn, cutting O′A′ at P′. Now O′P′ = OP = OQ.\n4. The compass is opened to the distance PQ: point on P, pencil on Q.\n5. Keeping that width, the compass point goes to P′ and cuts the long arc at Q′. Now P′Q′ = PQ.\n6. The ray O′Q′ is drawn. ∠A′O′Q′ is an exact copy of ∠AOB: a protractor reads 50° on both.\n\nWhy it works: triangle OQP has OQ = OP = r and QP = the chord. Triangle O′P′Q′ has O′P′ = O′Q′ = r (both on the arc of radius r, centre O′) and P′Q′ = the same chord. Three sides match, so the triangles are congruent by SSS, and the angles at O and at O′ are equal. A protractor placed on both angles gives the same reading.",{"id":513,"type":47,"variant":101,"title":514,"markdown":515},"mis-copy-radius","\"Any radius for the second arc will do\"","In the copying recipe, the arc at O′ **must** use the same radius as the arc at O. If it is wider, the chord PQ would sit on a bigger circle and cut off a **smaller** angle there, because a fixed chord on a larger circle subtends a smaller angle at the centre. The three-sides match only works if both arms match (r and r) as well as the chord.",{"id":517,"type":149,"itemId":518,"prompt":519,"check":520,"hints":531,"feedback":534},"pr-copy-why","constructing-angles.dee-copy-why","When you copy ∠AOB to O′, which congruence rule guarantees the copied angle is equal?",{"kind":153,"options":521,"correct":530},[522,524,526,528],{"id":156,"label":523},"SSS: two radii and the chord match",{"id":159,"label":525},"Angles on a straight line",{"id":162,"label":527},"The average rule",{"id":165,"label":529},"Measuring with a protractor",[156],[532,533],"List the three sides of each triangle.","OP = O′P′, OQ = O′Q′, PQ = P′Q′.",{"correct":535,"incorrect":536},"Right: OP = O′P′, OQ = O′Q′ (same radius) and PQ = P′Q′ (same chord), so SSS gives congruent triangles and equal angles.","The copy uses two equal radii and one equal chord, so the triangles have three equal sides: SSS congruence. A protractor is only used to *check*.",{"id":538,"type":47,"variant":67,"title":539,"markdown":540},"example-carpenter","Copying angles in a workshop","A carpenter fitting a shelf into an old wall corner that is not quite 90° uses a **sliding bevel**: a handle with a blade that locks at any angle. She sets it against the wall, locks it, and transfers the angle to the plank. It is exactly the copy-angle idea: capture the opening without ever turning it into a number, so no reading error creeps in.",{"id":542,"type":53,"title":543,"eyebrow":544,"navLabel":545},"ch9","Reflex angles and the limits of accuracy","Chapter 09","9 Reflex and accuracy",{"id":547,"type":43,"markdown":548},"reflex-prose","A protractor only goes up to 180°, so a **reflex angle** (between 180° and 360°) needs a little reasoning. There are two exact methods, and a careful measurer uses one to check the other.\n\n**Method 1, the leftover.** Measure the ordinary angle on the *other* side, call it x. The two angles together make a full turn, so the reflex angle is **360° − x**.\n\n**Method 2, straight line plus extra.** Extend one arm backwards through the vertex to make a straight line (180°). Measure the part of the reflex angle beyond that line, call it y. The reflex angle is **180° + y**.\n\nIf the two answers differ by more than a degree or two, one measurement is wrong.",{"id":550,"type":114,"title":551,"problem":552,"steps":553},"we-reflex","Measuring a reflex angle two ways","A reflex angle is drawn. Its non-reflex partner measures 125°. When one arm is extended, the part beyond the straight line measures 55°. Find the reflex angle both ways.",[554,555,556],"Method 1: 360° − 125° = **235°**.","Method 2: 180° + 55° = **235°**.","They agree. ✓ Notice why: 125° + 55° = 180°, because the non-reflex angle and the extra part together fill the straight line.",{"id":558,"type":559,"component":560,"componentVersion":5,"config":561,"objective":568,"textAlternative":569,"help":570},"lab-reflex","interactive","protractor",{"mode":81,"targets":562,"tolerance":5},[563,564,565,566,567],200,235,270,305,340,"Measure five reflex angles with a virtual protractor to within 1°, using 360° − x or 180° + y.","The lab draws five reflex angles, one at a time, with a virtual protractor you can place and rotate. The target accuracy is 1°.\n\n- **200°**: the other side measures 160°, and 360 − 160 = 200. Or extend an arm: the extra beyond 180° is 20°.\n- **235°**: the other side is 125°; 360 − 125 = 235. Extra beyond the straight line: 55°.\n- **270°**: the other side is a right angle, 90°; 360 − 90 = 270. Extra: exactly 90°, so both methods use a right angle.\n- **305°**: the other side is 55°; 360 − 55 = 305. Extra: 125°.\n- **340°**: the other side is only 20°; 360 − 20 = 340. Extra: 160°.\n\nFor each angle, first decide roughly: just past a straight line (about 200°), about three-quarters of a turn (270°) or nearly a full turn (340°). Then measure and subtract. A reading that disagrees with your estimate by tens of degrees usually means the wrong scale was used on the protractor.",{"simplerExplanation":571,"hints":572},"Measure the small angle outside instead, then take it away from 360°.",[573,574],"Measure the angle that is **not** shaded, then subtract from 360°.","If the small angle is 55°, the reflex angle is 360 − 55 = 305°.",{"id":576,"type":43,"markdown":577},"accuracy-prose","How accurate can a protractor measurement be? Two things limit it.\n\n**1. How far apart the degree marks are.** On a protractor of radius 5 cm, the arc for one degree is 2 × π × 5 ÷ 360 ≈ **0.87 mm** long. On a 10 cm board protractor it is about 1.75 mm. Your eye can reasonably split a gap of under a millimetre into halves at best, so ±1° is an honest target with a school protractor, and ±0.5° with a big one.\n\n**2. How well the arms are drawn and lined up.** Suppose the end of an arm is misplaced sideways by just **1 mm**. The angle error is the angle whose \"opposite over adjacent\" is 1 mm over the arm length. For a 4 cm arm that is about **1.43°**; for a 10 cm arm only about **0.57°**. Longer arms make the same slip matter less, which is exactly why teachers say: *extend short arms before you measure*.\n\nYou do not need trigonometry to believe the numbers below: draw a 10 cm arm, mark a point 1 mm to the side of its tip, join that point to the vertex and measure the new angle with a protractor. The scale drawing gives the same answer as the calculation.",{"id":579,"type":382,"caption":580,"columns":581,"rows":585},"t-errors","Angle error caused by a sideways slip at the end of an arm (computed with tan⁻¹)",[582,583,584],"Slip","Arm length","Angle error",[586,590,593,596,599],[587,588,589],"1 mm","2 cm","≈ 2.9°",[587,591,592],"4 cm","≈ 1.4°",[587,594,595],"10 cm","≈ 0.6°",[597,223,598],"2 mm","≈ 2.3°",[600,594,601],"0.5 mm (sharp pencil)","≈ 0.3°",{"id":603,"type":149,"itemId":604,"prompt":605,"check":606,"hints":609,"feedback":612},"pr-error","constructing-angles.dee-slip-error","The end of a 5 cm arm is drawn 2 mm to the side of where it should be. Roughly how many degrees is the angle off? Give your answer to one decimal place.",{"kind":490,"answer":607,"tolerance":608,"unit":493},2.3,0.2,[610,611],"The slip (0.2 cm) divided by the arm length (5 cm) is 0.04.","For small angles, the error in degrees is about 0.04 × 57.3.",{"correct":613,"incorrect":614},"Right: tan⁻¹(0.2 ÷ 5) ≈ 2.3°. That is more than the usual ±1° tolerance, so extend the arm or redraw it.","The ratio is 0.2 ÷ 5 = 0.04. The angle with tangent 0.04 is about 2.3° (a quick rule: ratio × 57.3 ≈ 2.3°).",{"id":616,"type":47,"variant":617,"title":618,"markdown":619},"model-limit","model_limit","What a protractor lab cannot show","The virtual protractor places its centre perfectly and draws infinitely thin lines, so its only error is where *you* choose to read. On paper there are more: the protractor's centre mark has width, the base line is printed with a thickness, the plastic may be slightly warped, and your eye may not be directly above the scale (this sideways-viewing error is called **parallax**). Real measurements are always \"value ± tolerance\", while a proof gives an exact value. Both are useful: the proof tells you what the answer *should* be, the measurement tells you how well you drew it.",{"id":621,"type":53,"title":622,"eyebrow":623,"navLabel":624},"ch10","The problems the Greeks could not crack","Chapter 10","10 Impossible problems",{"id":626,"type":43,"markdown":627},"three-problems","Bisecting any angle is easy. So the Greeks naturally asked: **can every angle be split into three equal parts (trisected)** with compass and straightedge alone? They also asked two other famous questions:\n\n- **Doubling the cube:** construct the edge of a cube with exactly twice the volume of a given cube.\n- **Squaring the circle:** construct a square with exactly the same area as a given circle.\n\nFor over two thousand years, brilliant mathematicians failed to find constructions. Many found clever methods that *cheated* slightly: Archimedes trisected any angle using a straightedge with **two marks** on it, slid into position (a move called *neusis*), which the strict rules forbid.\n\nIn **1837** the French mathematician **Pierre Wantzel** proved that trisecting a general angle and doubling the cube are **impossible** with compass and straightedge, and in **1882** Ferdinand von Lindemann's work on π showed that squaring the circle is impossible too. The key idea: every compass-and-straightedge step can only produce lengths built from whole numbers using +, −, ×, ÷ and square roots, and the lengths needed for those problems are not of that kind.",{"id":629,"type":264,"prompt":630,"options":631,"explanation":640},"pred-trisect90","Wantzel proved you cannot trisect *every* angle with compass and straightedge. So can you trisect a **90°** angle exactly?",[632,634,636,638],{"id":156,"label":633},"No: trisection is impossible",{"id":159,"label":635},"Yes: 90° ÷ 3 = 30°, and 30° is constructible",{"id":162,"label":637},"Only approximately",{"id":165,"label":639},"Only with a marked ruler","**Yes.** \"Impossible in general\" does not mean \"impossible for every angle\". One third of 90° is 30°, and you can construct 30° by bisecting 60°. The same goes for 180° (one third is 60°) and 45° (one third is 15°). What Wantzel proved is that there is **no single recipe that works for every angle**, and that some particular angles, such as **60°**, cannot be trisected: one third of 60° is 20°, and 20° cannot be constructed exactly.",{"id":642,"type":47,"variant":62,"title":643,"markdown":644},"nuance-impossible","\"Impossible\" is a proof, not a shrug","In everyday speech \"impossible\" often means \"nobody has managed it yet\". In mathematics it is a proven fact, like \"no odd number is divisible by 2\". People still post \"trisection constructions\" online; every one of them either breaks the rules (a marked ruler, guessing a point) or is only approximately right. The approximations can be astonishingly close, within a thousandth of a degree, but close is not exact. You will look at why 20° is out of reach, and at which regular polygons *can* be constructed, in the Extend layer.",{"id":646,"type":646,"title":647,"items":648},"timeline","Two and a half thousand years of compass and straightedge",[649,653,657,661,665,669,673,677,681],{"time":650,"title":651,"text":652},"Ancient","Babylonian degrees","Babylonian astronomers counted in base 60. The division of a full turn into 360 parts is usually traced to them and to later Greek astronomers.",{"time":654,"title":655,"text":656},"800–500 BCE","Sulba Sutras","Indian priests lay out fire altars with ropes and pegs: straight lines, circles, squares and right angles made from stretched cords.",{"time":658,"title":659,"text":660},"c. 300 BCE","Euclid's Elements","Book I opens by constructing an equilateral triangle, then shows how to bisect an angle, bisect a segment and draw perpendiculars.",{"time":662,"title":663,"text":664},"c. 250 BCE","Archimedes' trick","Archimedes trisects any angle with a straightedge carrying two marks, a move outside Euclid's rules.",{"time":666,"title":667,"text":668},"1672","Mohr","Georg Mohr's Euclides Danicus shows that every compass-and-straightedge point can be found with a compass alone. The book is then forgotten until 1928.",{"time":670,"title":671,"text":672},"1796","Gauss's 17-gon","Eighteen-year-old Carl Friedrich Gauss shows that a regular 17-sided polygon can be constructed, the first new polygon since the Greeks.",{"time":674,"title":675,"text":676},"1797","Mascheroni","Lorenzo Mascheroni independently proves the compass-only result; it becomes the Mohr–Mascheroni theorem.",{"time":678,"title":679,"text":680},"1837","Wantzel","Pierre Wantzel proves that trisecting a general angle and doubling the cube are impossible with compass and straightedge.",{"time":682,"title":683,"text":684},"1882","Lindemann","Ferdinand von Lindemann proves π is transcendental, so squaring the circle is impossible.",{"id":686,"type":47,"variant":128,"title":687,"markdown":688},"aha-compass-only","You do not even need the straightedge","The Mohr–Mascheroni theorem says that any point you can find with compass and straightedge, you can find with a **compass alone**. You cannot draw the straight line itself, of course, but you can find every point on it that matters. So the straightedge is, in a sense, a convenience. The compass is the truly powerful tool.",{"id":690,"type":53,"title":691,"eyebrow":692,"navLabel":693},"ch11","Pulling the reasons together","Chapter 11","11 Wrap-up",{"id":695,"type":559,"component":696,"componentVersion":5,"config":697,"objective":725,"textAlternative":726,"help":727},"lab-match","match-pairs",{"prompt":698,"mode":699,"pairs":700},"Match each construction to the reason it works.","connect",[701,704,707,710,713,716,719,722],{"a":702,"b":703},"60° construction","Triangle OPQ is equilateral",{"a":705,"b":706},"Angle bisector","Triangles OPT and OQT are congruent by SSS",{"a":708,"b":709},"Perpendicular bisector","P and Q are both equidistant from A and B",{"a":711,"b":712},"90° from the 60° and 120° marks","The bisector sits at the average, (60 + 120) ÷ 2",{"a":714,"b":715},"Copying an angle","Same radius and same chord, so SSS",{"a":717,"b":718},"Six compass steps round a circle","6 × 60° = 360°, a full turn",{"a":720,"b":721},"Perpendicular at a point P","Z is equidistant from X and Y, and P is their midpoint",{"a":723,"b":724},"Reflex angle from its partner","The two angles make a full turn: 360° − x","Connect each construction with the geometric fact that makes it exact.","Eight pairs to connect.\n\n- The **60° construction** works because triangle OPQ is **equilateral** (all sides are the one radius).\n- The **angle bisector** works because triangles OPT and OQT are **congruent by SSS**.\n- The **perpendicular bisector** works because P and Q are each **equidistant from A and B**.\n- **90° from the 60° and 120° marks** works because the bisector points to the **average**, (60 + 120) ÷ 2 = 90.\n- **Copying an angle** works because the **same radius and same chord** give congruent triangles (SSS).\n- **Six compass steps** close up because **6 × 60° = 360°**.\n- The **perpendicular at a point P** works because Z is equidistant from X and Y, and P is the midpoint of XY.\n- A **reflex angle** equals **360° − x** because it and its partner make a full turn.",{"hints":728},[729,730],"Look for the words equilateral, congruent, equidistant and average.","Which construction uses a chord copied from the original angle?",{"id":732,"type":559,"component":733,"componentVersion":5,"config":734,"objective":795,"textAlternative":796,"help":797},"lab-sort","sort-game",{"prompt":735,"bins":736,"items":746,"seconds":492},"Always, sometimes or never true? Sort each statement about constructions.",[737,740,743],{"id":738,"label":739},"always","Always true",{"id":741,"label":742},"sometimes","Sometimes true",{"id":744,"label":745},"never","Never true",[747,751,755,759,763,767,771,775,779,783,787,791],{"id":748,"label":749,"bin":738,"why":750},"s1","A point on the perpendicular bisector of AB is the same distance from A and from B.","That is the defining property: the perpendicular bisector is the locus of points equidistant from A and B.",{"id":752,"label":753,"bin":741,"why":754},"s2","In the angle-bisector recipe, the arcs from P and Q meet whatever radius you choose.","They meet only if the radius is at least half of PQ. Too small and they never cross.",{"id":756,"label":757,"bin":738,"why":758},"s3","Walking the radius round a circle six times brings you back exactly to the start.","Each step makes an equilateral triangle and turns 60°; 6 × 60° = 360°. In exact geometry it always closes.",{"id":760,"label":761,"bin":741,"why":762},"s4","Bisecting an angle gives two acute angles.","Bisect 70° and you get 35° + 35° (acute). Bisect 180° and you get two right angles; bisect a 300° reflex angle and you get two 150° angles.",{"id":764,"label":765,"bin":744,"why":766},"s5","A compass-and-straightedge construction gives exactly 20°.","20° is one third of 60°. Wantzel's work shows it cannot be constructed exactly.",{"id":768,"label":769,"bin":738,"why":770},"s6","The bisectors of the two angles in a linear pair meet at 90°.","x\u002F2 + (180 − x)\u002F2 = 90, whatever x is.",{"id":772,"label":773,"bin":744,"why":774},"s7","Changing the radius of the first arc changes the 60° angle you construct.","Any radius gives an equilateral triangle, so the angle is always exactly 60°.",{"id":776,"label":777,"bin":741,"why":778},"s8","Bisecting a whole number of degrees gives a whole number of degrees.","Bisect 60° and you get 30°; bisect 45° and you get 22.5°.",{"id":780,"label":781,"bin":741,"why":782},"s9","An angle can be trisected exactly with compass and straightedge.","90° can (into 30° parts) and 180° can (into 60° parts), but 60° cannot, since 20° is not constructible.",{"id":784,"label":785,"bin":744,"why":786},"s10","Copying an angle needs a protractor.","Two equal radii and a copied chord do it exactly; a protractor is only for checking.",{"id":788,"label":789,"bin":744,"why":790},"s11","A slanted line from a point to a road is shorter than the perpendicular.","The perpendicular is always the shortest; any slanted line is the hypotenuse of a right-angled triangle.",{"id":792,"label":793,"bin":741,"why":794},"s12","Two arcs of equal radius from A and B cross on the perpendicular bisector of AB.","When they cross at all, the crossing points are on the bisector; but if the radius is less than half of AB they do not cross.","Sort statements about constructions into always, sometimes and never true, and justify each.","Twelve statements to sort.\n\n**Always true:** points on the perpendicular bisector of AB are equidistant from A and B; six radius-steps close a circle (6 × 60° = 360°); the bisectors of a linear pair meet at 90° because x\u002F2 + (180 − x)\u002F2 = 90.\n\n**Sometimes true:** the arcs in the bisector recipe meet (only if the radius is at least half of PQ); bisecting gives two acute angles (not for 180° or reflex angles); bisecting a whole number of degrees gives a whole number (60° yes, 45° gives 22.5°); an angle can be trisected (90° yes, 60° no); equal arcs from A and B cross on the bisector (only if they cross at all).\n\n**Never true:** a construction gives exactly 20°; changing the first radius changes the 60° angle; copying an angle needs a protractor; a slanted line to a road is shorter than the perpendicular.",{"hints":798},[799,800],"For \"sometimes\", find one example where it works and one where it fails.","Think about the edge cases: tiny radii, straight angles, reflex angles.",{"id":802,"type":803,"prompt":804},"reflect","reflection","A classmate says: \"I measured my constructed angle and got 59°, so the construction must be wrong.\" Write a short reply explaining the difference between the construction being exact and the drawing being accurate, and suggest two things they could do to get closer to 60°.",{"id":806,"type":806,"title":807,"terms":808},"glossary","Words for reasoning about constructions",[809,813,817,821,825,829,833,837,841,845,849,853,857,861,865,867,870],{"term":810,"meaning":811,"example":812},"Congruent","Exactly the same shape and size: one figure fits on the other after sliding, turning or flipping.","Triangles OPT and OQT in the bisector construction.",{"term":814,"meaning":815,"example":816},"SSS rule","If all three sides of one triangle equal the three sides of another, the triangles are congruent.","Used to prove the angle bisector and the copy-angle constructions.",{"term":818,"meaning":819,"example":820},"SAS rule","If two sides and the angle between them match, the triangles are congruent.","Used in the kite proof of the perpendicular bisector.",{"term":822,"meaning":823,"example":824},"Equilateral triangle","A triangle with all three sides equal; each of its angles is 60°.","Triangle OPQ in the 60° construction.",{"term":826,"meaning":827,"example":828},"Isosceles triangle","A triangle with two equal sides; the angles opposite them are equal.","Triangle OPQ, with OP = OQ.",{"term":830,"meaning":831,"example":832},"Equidistant","The same distance from two (or more) points or lines.","P is equidistant from A and B.",{"term":834,"meaning":835,"example":836},"Locus","The set of all points that satisfy a rule.","The locus of points equidistant from A and B is the perpendicular bisector of AB.",{"term":838,"meaning":839,"example":840},"Rhombus","A four-sided shape with four equal sides; its diagonals bisect each other at right angles.","APBQ when the same radius is used above and below AB.",{"term":842,"meaning":843,"example":844},"Kite","A four-sided shape with two pairs of equal adjacent sides; one diagonal is the perpendicular bisector of the other.","APBQ when different radii are used above and below AB.",{"term":846,"meaning":847,"example":848},"Chord","A straight segment joining two points on a circle.","PQ in the copy-angle construction.",{"term":850,"meaning":851,"example":852},"Straightedge","A tool for drawing straight lines, with no measuring marks used.","A ruler used without reading its numbers.",{"term":854,"meaning":855,"example":856},"Collapsing compass","Euclid's imagined compass that snaps shut when lifted, so it cannot carry a distance.","Euclid proved it can still copy lengths (Elements I.2).",{"term":858,"meaning":859,"example":860},"Trisect","Divide into three equal parts.","Trisecting 90° gives three 30° angles.",{"term":862,"meaning":863,"example":864},"Neusis","A construction that slides a marked ruler into place; not allowed by Euclid's rules.","Archimedes' method for trisecting any angle.",{"term":659,"meaning":866},"A Greek geometry textbook from about 300 BCE that builds geometry from a few assumptions using constructions and proofs.",{"term":868,"meaning":869},"Parallax","The reading error caused by looking at a scale from the side instead of straight down.",{"term":871,"meaning":872,"example":873},"Tolerance","How far a measurement may be from the true value and still be accepted.","±1° for a school protractor.",{"id":875,"type":875,"title":9,"questions":876},"quiz",[877,890,903,916,926,939,952,965,978,991,1004,1017],{"itemId":878,"prompt":879,"options":880,"correct":156,"why":889},"constructing-angles.dee-q-sixty","In the 60° construction, which three lengths are equal?",[881,883,885,887],{"id":156,"label":882},"OP, OQ and PQ",{"id":159,"label":884},"OA, OP and OQ",{"id":162,"label":886},"OP, PQ and QA",{"id":165,"label":888},"Only OP and OQ","OP and OQ are radii of the first arc; PQ is the radius of the second arc, which is the same. So triangle OPQ is equilateral and ∠QOP = 60°.",{"itemId":891,"prompt":892,"options":893,"correct":156,"why":902},"constructing-angles.dee-q-sss","Which rule proves that the angle-bisector construction works?",[894,896,898,900],{"id":156,"label":895},"SSS congruence",{"id":159,"label":897},"Angles on a straight line add to 180°",{"id":162,"label":899},"Vertically opposite angles",{"id":165,"label":901},"The angles of a quadrilateral add to 360°","OP = OQ, PT = QT and OT is shared, so triangles OPT and OQT are congruent by SSS, giving ∠POT = ∠QOT.",{"itemId":904,"prompt":905,"options":906,"correct":156,"why":915},"constructing-angles.dee-q-locus","Point X is 5 cm from A and 5 cm from B. Where must X be?",[907,909,911,913],{"id":156,"label":908},"On the perpendicular bisector of AB",{"id":159,"label":910},"On the segment AB",{"id":162,"label":912},"At the midpoint of AB",{"id":165,"label":914},"Anywhere 5 cm from AB","Any point equidistant from A and B lies on the perpendicular bisector of AB. X could be the midpoint only if AB = 10 cm.",{"itemId":917,"prompt":918,"options":919,"correct":156,"why":925},"constructing-angles.dee-q-arcs","You bisect a 60° angle whose first arc has radius 8 cm. The arcs from P and Q need a radius of more than:",[920,921,923,924],{"id":156,"label":591},{"id":159,"label":922},"8 cm",{"id":162,"label":588},{"id":165,"label":225},"For a 60° angle, triangle OPQ is equilateral, so PQ = 8 cm. The arcs must be more than half of PQ, which is 4 cm.",{"itemId":927,"prompt":928,"options":929,"correct":156,"why":938},"constructing-angles.dee-q-ninety","Rays at 60° and 120° are bisected. Why is the bisector at exactly 90°?",[930,932,934,936],{"id":156,"label":931},"It sits at the average: (60 + 120) ÷ 2",{"id":159,"label":933},"Because 120 − 60 = 60",{"id":162,"label":935},"Because 60 + 120 = 180 is a straight line",{"id":165,"label":937},"Because the compass radius was 4 cm","The gap is 60° wide, so the bisector is 30° past the 60° ray: 60 + 30 = 90, which equals (60 + 120) ÷ 2.",{"itemId":940,"prompt":941,"options":942,"correct":156,"why":951},"constructing-angles.dee-q-105","Which pair of rays should you bisect between to get 105°?",[943,945,947,949],{"id":156,"label":944},"90° and 120°",{"id":159,"label":946},"60° and 135°",{"id":162,"label":948},"45° and 60°",{"id":165,"label":950},"90° and 180°","(90 + 120) ÷ 2 = 105. The other pairs give (60 + 135) ÷ 2 = 97.5°, (45 + 60) ÷ 2 = 52.5° and (90 + 180) ÷ 2 = 135°.",{"itemId":953,"prompt":954,"options":955,"correct":156,"why":964},"constructing-angles.dee-q-hexagon","Why does walking the radius round a circle make a regular hexagon?",[956,958,960,962],{"id":156,"label":957},"Each step turns 60° at the centre and 6 × 60° = 360°",{"id":159,"label":959},"A hexagon has 6 sides so it always fits",{"id":162,"label":961},"The radius is exactly 6 cm",{"id":165,"label":963},"Because 360 ÷ 60 = 5","Each chord equal to the radius forms an equilateral triangle with the centre, so each step turns 60°. Six steps make a full 360° turn and land back at the start.",{"itemId":966,"prompt":967,"options":968,"correct":156,"why":977},"constructing-angles.dee-q-copy","When copying an angle, the compass is set to the distance PQ. Why?",[969,971,973,975],{"id":156,"label":970},"So the new triangle has the same third side (chord) as the original",{"id":159,"label":972},"To measure the angle in centimetres",{"id":162,"label":974},"To make the arms longer",{"id":165,"label":976},"To find the midpoint of the angle","O′P′ = OP and O′Q′ = OQ already (the same radius). Copying the chord PQ across to P′Q′ gives the third matching side, so SSS makes the angles equal.",{"itemId":979,"prompt":980,"options":981,"correct":156,"why":990},"constructing-angles.dee-q-reflex","The non-reflex side of an angle measures 38°. What is the reflex angle?",[982,984,986,988],{"id":156,"label":983},"322°",{"id":159,"label":985},"142°",{"id":162,"label":987},"218°",{"id":165,"label":989},"342°","The two angles make a full turn: 360° − 38° = 322°. (142° is the supplement; 218° is 180° + 38°, a common slip.)",{"itemId":992,"prompt":993,"options":994,"correct":156,"why":1003},"constructing-angles.dee-q-slip","A 1 mm slip at the end of an arm causes a smaller angle error when the arm is:",[995,997,999,1001],{"id":156,"label":996},"Longer",{"id":159,"label":998},"Shorter",{"id":162,"label":1000},"Thicker",{"id":165,"label":1002},"Drawn in pen","The error is roughly slip ÷ arm length (in radians). A 1 mm slip is about 1.4° on a 4 cm arm but only about 0.6° on a 10 cm arm.",{"itemId":1005,"prompt":1006,"options":1007,"correct":156,"why":1016},"constructing-angles.dee-q-trisect","Which of these angles can be trisected exactly with compass and straightedge?",[1008,1010,1012,1014],{"id":156,"label":1009},"90°",{"id":159,"label":1011},"60°",{"id":162,"label":1013},"Every angle",{"id":165,"label":1015},"No angle","One third of 90° is 30°, which is constructible. One third of 60° is 20°, which is not. Wantzel (1837) proved there is no method that works for every angle.",{"itemId":1018,"prompt":1019,"options":1020,"correct":156,"why":1029},"constructing-angles.dee-q-compass-only","The Mohr–Mascheroni theorem says:",[1021,1023,1025,1027],{"id":156,"label":1022},"Every constructible point can be found with a compass alone",{"id":159,"label":1024},"Every angle can be trisected",{"id":162,"label":1026},"A straightedge alone can bisect angles",{"id":165,"label":1028},"A protractor is more exact than a compass","Mohr (1672) and Mascheroni (1797) showed that the straightedge is not needed to find points: the compass alone can locate them all.",{"id":1031,"type":1032,"title":1033,"points":1034},"cheat","summary","Cheat sheet",[1035,1036,1037,1038,1039,1040,1041,1042,1043,1044,1045],"**Compass and straightedge only:** new points come only from where lines and circles cross, so a construction plus its reason is exact for any size of drawing.","**60°:** OP = OQ = PQ (one radius), so triangle OPQ is equilateral and each angle is 180° ÷ 3 = 60°. Six steps round a circle close because 6 × 60° = 360°.","**Angle bisector:** OP = OQ, PT = QT, OT shared → triangles congruent (SSS) → equal halves. The arcs from P and Q must have radius more than half of PQ.","**Perpendicular bisector:** it is the locus of points equidistant from A and B. Equal radii make a rhombus (different radii a kite); its diagonals cross at right angles.","**Perpendiculars at and from a point** are the perpendicular bisector idea in disguise. The perpendicular distance is the shortest distance to a line.","**Average rule:** bisecting between rays at a° and b° gives (a + b) ÷ 2. So 90 = (60 + 120) ÷ 2, 75 = (60 + 90) ÷ 2, 105 = (90 + 120) ÷ 2, 135 = (90 + 180) ÷ 2.","**Reachable with 60° steps, bisection and straight lines:** every multiple of 15° up to 180°, and their halves (7.5°, 22.5°, 37.5°…). Never 20°, 40°, 50°, 70° or 80°.","**Copying an angle:** same radius at both vertices plus the same chord → SSS → equal angles.","**Reflex angles:** 360° − x (x = the other side) or 180° + y (y = the extra past a straight line). Use one method to check the other.","**Accuracy:** 1° is under 1 mm of arc on a 5 cm protractor. A 1 mm slip is ≈ 1.4° on a 4 cm arm but ≈ 0.6° on a 10 cm arm, so extend short arms.","**Impossible problems:** a general angle cannot be trisected (Wantzel, 1837), so 20° cannot be constructed; but special angles like 90° and 180° can be trisected.",{"id":1047,"type":1048,"conceptId":1049,"relation":1050,"explanation":1051},"conn-angles","connection","angles","helps_understand","Angles on a straight line, angle sums in triangles and linear pairs are the facts that turn each construction recipe into a proof.",{"id":1053,"type":1048,"conceptId":1054,"relation":1055,"explanation":1056},"conn-shape","shape-and-space","applied_in","Equilateral triangles, rhombuses, kites and the regular hexagon all appear inside these constructions, and constructions let you draw those shapes exactly.",{"id":1058,"type":1048,"conceptId":1059,"relation":1060,"explanation":1061},"conn-lines","lines","related_to","Perpendicular lines, perpendicular bisectors and the shortest distance from a point to a line all rest on the constructions in this layer.",{"id":1063,"type":1063,"sourceIds":1064},"sources",[1065,1066,1067,1068,1069,1070,1071,1072,1073,1074,1075,1076,1077],"constructing-angles-ncert-math-6-practical-geometry","constructing-angles-ncert-ganita-prakash-6","constructing-angles-ncert-math-7-practical-geometry","constructing-angles-mathsisfun-protractor","constructing-angles-mathsisfun-constructions","constructing-angles-mathsisfun-degrees","constructing-angles-wikipedia-straightedge-compass","constructing-angles-wikipedia-angle-trisection","constructing-angles-wikipedia-shulba-sutras","constructing-angles-britannica-euclid-elements","constructing-angles-sciam-heptadecagon","constructing-angles-wikipedia-mohr-mascheroni","constructing-angles-wikipedia-squaring-circle",[1065,1066,1067,1068,1069,1070,1071,1072,1073,1074,1075,1076,1077],"needs_review",{"generatedBy":1081,"notes":1082},"claude-code","Draft generated with Python; every angle fact was computed and asserted. Pending owner review.","2fe9e7afa5551547d93a535637fe94e4d6f52c90d9ba0bc887b155057622eac9",{"logic:practice":1085,"component:compass-construction@1":1086,"component:protractor@1":1087,"component:match-pairs@1":1088,"component:sort-game@1":1089,"source:constructing-angles-britannica-euclid-elements":1090,"source:constructing-angles-mathsisfun-constructions":1091,"source:constructing-angles-mathsisfun-degrees":1092,"source:constructing-angles-mathsisfun-protractor":1093,"source:constructing-angles-ncert-ganita-prakash-6":1094,"source:constructing-angles-ncert-math-6-practical-geometry":1095,"source:constructing-angles-ncert-math-7-practical-geometry":1096,"source:constructing-angles-sciam-heptadecagon":1097,"source:constructing-angles-wikipedia-angle-trisection":1098,"source:constructing-angles-wikipedia-mohr-mascheroni":1099,"source:constructing-angles-wikipedia-shulba-sutras":1100,"source:constructing-angles-wikipedia-squaring-circle":1101,"source:constructing-angles-wikipedia-straightedge-compass":1102},"3d6b0fe1b15255975a32b0fcd94e8019bc959ad45cbf12e136e86149549c6878","0eba81381f31d8d78008911eb4c3d62745efd33b4ffadd94aef0851d0a2ad5f3","7407db21456592711dd16c6bdad23f042e85ebab9c314b072b17b7065eaf8ae3","2a8ee4ac87460b4e1175a4bb13c96b03d577db06dde95670eb7fcfe4ad787899","b164f45a2c8ca08f26c450768ff0231e113e9fe45381eddb34dc6d0548596c38","ad1a9a6a227fda5d3c1569f37efbe35e448ebaceba8cba872821fd48e2e00ed6","b93faffbd9c4d40f5fce2bc4b2ea0ab5ac64bb8c176f5e2bba3f37444df5e400","216eb0db510461864a47157f14054a39e15b1b0fc461b0fbc77664d9eb28b91d","4d3f50c07f44df57c80455dc39e01b2aa11bb0ee40811fca3b0f12d16e0b3f5e","acad5a4d56d24a5c1ad6e908f3809f2e7b3978f7c2810a9b33cdf82a65c4bb47","4aedaa1be389589b6e840923ef4e92fd15d03eda0b0ba0302d57b7e71bcb3dc7","21119e12648b9efd4cc82b11c59d626f2a53eace3a70552041f26c311b77ba2d","2ab885aef682a4c817268dd75b6110889b7897fc18b98ebc081e4aa62b5419f7","7f00387dc29d17172d25b6aa96420e2544a8bc59edf939af3dce91d515300a1f","cc25677292f8107d22a9d8d355bac95ae2a897ec0b3e109994d783824029d895","f7559697a2f963f9cb1e02a93fc5697840f583f22605d7483d688664862d70f9","d66ae780e95eb677a3ae7bcb24aa2430951fe7dded13b8366b75578ca4089dee","90d6cc756bc1bdd6cde0d5e4ed2000c88c3e2f3a8d91fdaf0ec4ab73af72b434",{"state":1104,"reviewer":1105,"selfReview":1106,"reviewedAt":1107,"method":1108},"approved","The library owner",true,"2026-09-20T10:18:37.581Z","owner_bulk","preview-7e1cbbcc4f",1789899598415]