Chapter 10: Congruent Triangles

Two plane figures are congruent if they have the same size and shape. Such figures can always be made to coincide, so congruent figures are called exact duplicates of each other. When two figures A and B are congruent, it is denoted by A ≅ B.

The process of matching two figures is called superposition.

The angles and sides which would match are called corresponding or homologous parts.

A triangle has six elements, three sides and three angles. The six elements of triangle ABC are AB, BC, AC, ∠BAC or ∠A, ∠CBA or ∠B, and ∠ACB or ∠C.

Two triangles △ABC and △DEF are said to be congruent if the following six conditions all hold: ∠A ≅ ∠D, ∠B ≅ ∠E, ∠C ≅ ∠F, AB ≅ DE, BC ≅ EF, AC ≅ DF. In this case, we write △ABC ≅ △DEF.

△ABC ≅ △ABC (identity congruence), i.e. every triangle is congruent to itself. If △ABC ≅ △DEF then △DEF ≅ △ABC (symmetric property of congruence). If △ABC ≅ △DEF and △DEF ≅ △XYZ, then △ABC ≅ △XYZ (transitive property of congruence).

The angle formed by two sides of a triangle is called the angle included between those two sides.

The side joining the vertices of two angles is the side included between those angles; either of the other two sides is a non-included side.

A triangle with two congruent sides is called an isosceles triangle. The third side is the base, the angle opposite the base is the vertex angle, and the two angles at the base are the base angles.

In a right-angled triangle the side opposite the right angle is the hypotenuse; the two sides that form the right angle are the legs.

A triangle in which all the three sides and all the three angles are equal is called an equilateral triangle.

Two lines or segments are perpendicular if they meet at a right angle (90°). We write ⊥.

A ray that divides an angle into two congruent angles is the bisector of that angle.

The midpoint of a segment divides it into two congruent segments. A median of a triangle is the segment joining a vertex to the midpoint of the opposite side.

The perpendicular bisector of a segment is the line that passes through its midpoint and is perpendicular to it.

When two geometric figures are identical (they can be placed exactly on top of each other), the figures are called congruent. Congruent figures are exact duplicates of each other. They have exactly the same size and shape.

A triangle has six elements, three sides and three angles.

CPCTC means Corresponding parts of congruent triangles are congruent.

If two sides and their included angle in one triangle are congruent to two sides and their included angle in another triangle, then the two triangles are congruent.

If two angles and a non-included side of one triangle are congruent to the corresponding angles and non-included side of another triangle, then the triangles are congruent.

A triangle in which all the three sides and all the three angles are equal is called an equilateral triangle.

If two angles and their included side in one triangle are congruent to two angles and their included side in another triangle, then the two triangles are congruent.

If two angles of a triangle are congruent, then the sides opposite to those angles are congruent.

If the sides of one triangle are congruent to the sides of another triangle, then the two triangles are congruent.

If the hypotenuse and leg of a right triangle are congruent to the hypotenuse and corresponding leg of another right triangle, then the two triangles are congruent.

Every proof in Exercise 10.1 and the Review Exercise ends with one of these five congruence rules, then CPCTC. Learn the name, what it needs, and when to pick it. The rule is written the way you must quote it in a reason: “S.A.S ≅ S.A.S”.

If two sides and their included angle in one triangle are congruent to two sides and their included angle in another triangle, then the two triangles are congruent.

If two angles and a non-included side of one triangle are congruent to the corresponding angles and non-included side of another triangle, then the triangles are congruent.

If two angles and their included side in one triangle are congruent to two angles and their included side in another triangle, then the two triangles are congruent.

In a correspondence of two triangles, if three sides of one triangle are congruent to the corresponding three sides of the other, the two triangles are congruent.

If in the correspondence of two right-angled triangles, the hypotenuse and one side of one are respectively congruent to the hypotenuse and the corresponding side of the other, the triangles are congruent.

If two angles of a triangle are congruent, then the sides opposite to those angles are congruent.

Corresponding parts of congruent triangles are congruent.

Every line of a Statements / Reasons table needs a reason from this list — the same wording the textbook uses.

This is the theorem you used in Exercise 10.1 Question 1. The proof works backwards: it supposes the two triangles are not congruent and shows that this leads to something impossible.

1 Draw the two triangles — Draw △ABC and △DEF side by side and write the correspondence △ABC ↔ △DEF. The double arrow means “we are pairing them up”; it does not yet say they are congruent.

2 Mark what is given — Three things are given. Put a tick on BC and EF, one arc at B and E, two arcs at A and D. BC ≅ EF — one equal side. ∠B ≅ ∠E. ∠A ≅ ∠D. Since two angles match, the third must match too: ∠C ≅ ∠F, because the three angles of a triangle add up to 180°.

3 Make the supposition and construct D′ — Suppose AC and DF are not equal. Then somewhere on DF there must be a point D′ with D′F equal to AC. Mark it and join D′ to E with a dashed line. We will now show that this supposition forces D′ to sit exactly on D — which means AC and DF were equal all along.

4 Apply S.A.S to △ABC and △D′EF — Compare △ABC with the small triangle △D′EF: BC ≅ EF — given. ∠C ≅ ∠F — shown in step 2. AC ≅ D′F — by our supposition. Side, included angle, side. So △ABC ≅ △D′EF by S.A.S, and therefore ∠B ≅ ∠D′EF.

5 Reach the contradiction — But ∠B ≅ ∠DEF was given. So ∠D′EF ≅ ∠DEF — the ray ED and the ray D′E make the same angle at E. Two rays from E that make the same angle with EF must be the same ray, so ED ≅ ED′ and D′ is the point D. Our supposition has collapsed: AC ≅ DF after all.

6 Finish with S.A.S — Now in △ABC ↔ △DEF we have BC ≅ EF (given), AC ≅ DF (just proved) and ∠C ≅ ∠F. Side, included angle, side again. ∴ △ABC ≅ △DEF

7 Corollary — A.S.A — If two angles and their included side in one triangle are congruent to two angles and their included side in another triangle, then the two triangles are congruent.

8 Why it follows from A.A.S — In A.S.A the equal side joins the two corners whose angles are equal. Because two equal angles force the third angle to be equal as well, an A.S.A match is automatically an A.A.S match. That is why the book prints A.S.A as a corollary of Theorem 10.1 instead of proving it again.

9 Used in: Ex 10.1 Q1 · Ex 10.1 Q2 · Review Q2 · A.S.A: Ex 10.1 Q3 · Review Q3

GIVEN In △ABC ↔ △DEF; BC ≅ EF, ∠B ≅ ∠E and ∠A ≅ ∠D

TO PROVE △ABC ≅ △DEF

CONSTRUCTION Suppose AC is not congruent to DF. Take a point D′ on DF such that AC ≅ D′F. Join D′ to E.

STEP 1 In △ABC ↔ △D′EF

STEP 2 BC ≅ EF — (Given)

STEP 3 ∠C ≅ ∠F — (∠B ≅ ∠E and ∠A ≅ ∠D)

STEP 4 AC ≅ D′F — (Supposition)

STEP 5 ∴ △ABC ≅ △D′EF — (S.A.S ≅ S.A.S)

STEP 6 Hence ∠B ≅ ∠D′EF — (Corresponding angles of congruent △s)

STEP 7 But ∠B ≅ ∠DEF — (Given)

STEP 8 ∴ ∠D′EF ≅ ∠DEF — (Transitive property)

STEP 9 This implies ED ≅ ED′, and thus AC ≅ DF

STEP 10 In △ABC ↔ △DEF: BC ≅ EF, AC ≅ DF, ∠C ≅ ∠F — (Given, Proved, Given)

STEP 11 ∴ △ABC ≅ △DEF — (S.A.S ≅ S.A.S)

The original theorem says equal sides give equal base angles. This one runs the other way: equal angles give equal sides.

1 Draw the triangle and name it — Draw any triangle and call it ABC. Do not draw it with two equal sides — that is what we have to prove, not what we are given.

2 Mark the two equal angles — Shade ∠B and ∠C. This is the only thing the question gives us.

3 Construct the bisector of ∠A — Draw the bisector of ∠A. Let it meet BC at D. By the meaning of a bisector, ∠1 ≅ ∠2. The bisector splits the big triangle into two smaller ones: △ABD and △ACD.

4 Match three parts — In △ABD ↔ △ACD: AD ≅ AD — the same side belongs to both (common). ∠1 ≅ ∠2 — from the construction. ∠B ≅ ∠C — given. Angle, angle and a non-included side (AD does not lie between ∠1 and ∠B), so the rule is A.A.S.

5 Read off the equal sides — The triangles are congruent, so all corresponding parts match. AB in the first triangle corresponds to AC in the second. ∴ AB ≅ AC

6 Used in: Review Q4 · and its reverse inside the proof of Theorem 10.4

GIVEN In △ABC, ∠B ≅ ∠C

TO PROVE AB ≅ AC

CONSTRUCTION Draw the bisector of ∠A cutting BC at the point D.

STEP 1 In △ABD ↔ △ACD

STEP 2 AD ≅ AD — (Common)

STEP 3 ∠1 ≅ ∠2 — (Construction)

STEP 4 ∠B ≅ ∠C — (Given)

STEP 5 ∴ △ABD ≅ △ACD — (A.A.S ≅ A.A.S)

STEP 6 Hence AB ≅ AC — (Corresponding sides of congruent △s)

This is the longest proof of the unit. The idea: build a copy of △ABC on the other side of EF, then show the copy is the mirror image of △DEF.

1 Draw both triangles — Draw △ABC and △DEF. Only the sides are given here — no angle is given at all.

2 Mark the three pairs of equal sides — One tick for AB and DE, two ticks for BC and EF, three ticks for CA and FD. Different numbers of ticks keep the pairs apart.

3 Construct △GEF below EF — Build a copy of △ABC onto EF, but on the far side from D: take GE ≅ AB and ∠GEF = ∠B. By S.A.S this copy is congruent to △ABC. The figure now looks like a kite: D on top, G below, EF across the middle.

4 Join D to G and number the angles — Draw DG. It splits the kite into two isosceles triangles. Number the four angles it makes: ∠1 = ∠EDG, ∠3 = ∠FDG at D, and ∠2 = ∠EGD, ∠4 = ∠FGD at G.

5 Use the triangle FDG — GF ≅ AC (corresponding sides of the congruent copy) and DF ≅ AC (given), so GF ≅ DF by the transitive property. So △FDG has two equal sides — it is isosceles — and its base angles are equal: ∠3 ≅ ∠4.

6 Use the triangle EGD — DE and GE are each congruent to AB, so DE ≅ GE. That makes △EGD isosceles too, and ∠1 ≅ ∠2.

7 Add the angles — ∠1 ≅ ∠2 and ∠3 ≅ ∠4. Adding the two pairs gives ∠1 + ∠3 ≅ ∠2 + ∠4, that is ∠D ≅ ∠G (addition of angles postulate).

8 Finish with S.A.S and transitivity — In △DEF ↔ △GEF: DE ≅ GE, ∠D ≅ ∠G and DF ≅ GF. So △DEF ≅ △GEF by S.A.S. But △ABC ≅ △GEF by the construction. Two triangles congruent to the same triangle are congruent to each other. ∴ △ABC ≅ △DEF

9 The book adds a note: this theorem is quoted in proofs as S.S.S ≅ S.S.S.

10 Used in: Ex 10.1 Q4 · Review Q5

GIVEN In △ABC ↔ △DEF; AB ≅ DE, BC ≅ EF and CA ≅ FD

TO PROVE △ABC ≅ △DEF

CONSTRUCTION Suppose the side BC is not shorter than either of the other two sides. Construct △GEF with G and D on opposite sides of EF such that GE ≅ AB and ∠B = ∠GEF. Join G and D, and number the angles as in the figure.

STEP 1 In △ABC ↔ △GEF: BC ≅ EF — (Given)

STEP 2 ∠B ≅ ∠GEF — (Construction)

STEP 3 AB ≅ GE — (Construction)

STEP 4 ∴ △ABC ≅ △GEF — (S.A.S ≅ S.A.S)

STEP 5 So GF ≅ AC — (Corresponding sides of congruent △s)

STEP 6 But DF ≅ AC — (Given)

STEP 7 ∴ GF ≅ DF — (i) — (Transitive property)

STEP 8 In △FDG, ∠3 ≅ ∠4 — (ii) — (△DFG is isosceles)

STEP 9 In △EGD, DE ≅ GE — (each is congruent to AB)

STEP 10 ∴ ∠1 ≅ ∠2 — (iii)

STEP 11 ∴ ∠3 + ∠1 ≅ ∠4 + ∠2 — (By (ii) and (iii))

STEP 12 ∴ m∠D ≅ m∠G — (Addition of angles postulate)

STEP 13 In △DEF ↔ △GEF: DE ≅ GE, ∠D ≅ ∠G, DF ≅ GF — (Proved, Proved, By (i))

STEP 14 ∴ △DEF ≅ △GEF — (S.A.S ≅ S.A.S)

STEP 15 But △ABC ≅ △GEF — (Proved above)

STEP 16 ∴ △ABC ≅ △DEF — (Transitive property)

Two matches instead of three — because the right angle is doing the work of the third.

1 Draw the two right triangles — Draw △DEF with its right angle at E, and △ABC with its right angle at B. Put the small square at each right angle.

2 Mark the hypotenuse and the leg — AC and DF are the hypotenuses — each faces the right angle. Mark them equal. Mark the legs BC and EF equal too. These are the only two facts given.

3 Produce AB to G and join CG — Extend the line AB past B to a point G with BG ≅ DE, then join C to G. Because ∠ABC is a right angle and A, B, G lie on one straight line, ∠CBG is a right angle too (supplementary angles).

4 Match △GBC with △DEF — In △GBC ↔ △DEF: GB ≅ DE (construction), ∠GBC ≅ ∠DEF (each a right angle), BC ≅ EF (given). Side, included angle, side. So △GBC ≅ △DEF and hence GC ≅ DF. But DF ≅ AC was given, so GC ≅ AC.

5 Use the isosceles triangle CAG — GC ≅ AC makes △CAG isosceles, so the angles opposite those two sides are equal: ∠A ≅ ∠G.

6 Close the chain — In △ABC ↔ △GBC: AC ≅ GC (proved), ∠A ≅ ∠G (proved), ∠ABC ≅ ∠GBC (each a right angle). That is angle, angle and a non-included side, so △ABC ≅ △GBC. And △GBC ≅ △DEF from step 4. By the transitive property: ∴ △ABC ≅ △DEF

7 The book adds a note: this theorem is quoted in proofs as H.L ≅ H.L.

8 Used in: Ex 10.1 Q7

GIVEN In △ABC ↔ △DEF; ∠ABC ≅ ∠E (each is a right angle); AC ≅ DF and BC ≅ EF

TO PROVE △ABC ≅ △DEF

CONSTRUCTION Produce AB to the point G such that BG ≅ DE, then join C and G.

STEP 1 In △GBC, m∠CBG = 90° — (Supplementary angle of the right angle ∠ABC)

STEP 2 In △GBC ↔ △DEF: GB ≅ DE — (Construction)

STEP 3 ∠GBC ≅ ∠DEF — (Each is a right angle)

STEP 4 BC ≅ EF — (Given)

STEP 5 ∴ △GBC ≅ △DEF — (S.A.S ≅ S.A.S)

STEP 6 GC ≅ DF — (Corresponding sides of congruent △s)

STEP 7 But DF ≅ AC, so GC ≅ AC — (Given)

STEP 8 It follows that ∠A ≅ ∠G — (Angles opposite the congruent sides of △CAG)

STEP 9 In △ABC ↔ △GBC: AC ≅ GC, ∠A ≅ ∠G, ∠ABC ≅ ∠GBC — (Proved, Proved, Each a right angle)

STEP 10 ∴ △ABC ≅ △GBC — (A.A.S ≅ A.A.S — printed as S.A.S in the book)

STEP 11 But △GBC ≅ △DEF — (Proved)

STEP 12 ∴ △ABC ≅ △DEF — (Transitive property)

Every word in this question is explained in Part 1 above. This is the question the whole method was built on, so the steps are shown in full.

1 Draw the isosceles triangle and name it — Draw a triangle with the pointed corner at the top. Name the top corner A, the bottom-left B and the bottom-right C. The two slanting sides AB and AC are equal — put one tick on each. The bottom side BC is the base.

2 Find the endpoints of the base — The base BC has two ends: the point B and the point C. The question talks about these two points only. From each of these points we are going to draw one perpendicular.

3 Draw the first perpendicular, from B — Which side is opposite to B? The side that B does not touch — that is AC. From B draw a line that meets AC at 90°. Name the point where it lands D. Draw the small square at D to show the right angle. Now BD ⊥ AC.

4 Draw the second perpendicular, from C — Do the same from the other endpoint. The side opposite to C is AB. From C draw a line meeting AB at 90°, and name the landing point E. Now CE ⊥ AB. We now have the two perpendiculars the question is talking about: BD and CE.

5 Say clearly what we have to prove — The question ends with “…are congruent”. It is talking about the two perpendiculars. So the whole job is to show: BD ≅ CE; In words: the blue segment and the red segment have the same length.

6 Write it the textbook way: Given and To Prove — Before any proof, turn the sentence of the question into maths. This is what the examiner wants to see first. Given: In △ABC, AB ≅ AC; BD ⊥ AC and CE ⊥ AB; To prove: BD ≅ CE; No construction is needed in this question — the perpendiculars are already given to us in the words of the question.

7 Find two triangles that hold BD and CE — Here is the trick used in almost every question of this chapter: to prove two segments equal, find a triangle around each one and prove the two triangles congruent. BD is a side of △ABD. CE is a side of △ACE. Pull the two triangles out of the big figure and look at them separately.

8 Compare the two triangles, part by part — △ACE has been flipped over like a page, so that it sits the same way as △ABD. Now the matching parts are easy to see. Three things are equal: ∠A ≅ ∠A — both triangles share the very same corner at A. We say it is common. ∠ADB ≅ ∠AEC — each is 90°, because BD and CE are perpendiculars. AB ≅ AC — this was given in the question (the two ticked sides).

9 Pick the rule: A.A.S — Count what we matched: an angle (∠A), an angle (90°) and a side (AB with AC). Is the side between the two angles? No — the side between ∠A and ∠D would be AD, and we know nothing about AD. Our side AB is not included, so the rule is A.A.S (Theorem 10.1). ∴ △ABD ≅ △ACE; Keep the letters in matching order: A ↔ A, B ↔ C, D ↔ E. The order tells the reader which part matches which.

10 Finish: corresponding sides are congruent — The triangles are congruent, so every matching part is equal (CPCTC). Which side matches BD? In △ABD, BD is the side opposite ∠A. In △ACE, the side opposite ∠A is CE. They correspond. ∴ BD ≅ CE; Put two ticks on each of them in the figure. This is exactly what the question asked, so the proof is complete.

11 How to attack any question like this — (1) Draw the figure exactly as the words say, and name every point. (2) Write Given and To prove in maths symbols. (3) Find a triangle around each of the two segments you must prove equal. (4) Match three parts of those triangles and name the rule — S.A.S, A.A.S, S.S.S or H.L. (5) Finish with “corresponding sides of congruent triangles”.

GIVEN In △ABC, AB ≅ AC; BD ⊥ AC and CE ⊥ AB

TO PROVE BD ≅ CE

STEP 1 In △ABD ↔ △ACE

STEP 2 ∠A ≅ ∠A — (Common)

STEP 3 ∠ADB ≅ ∠AEC — (Each is a right angle, BD ⊥ AC and CE ⊥ AB)

STEP 4 AB ≅ AC — (Given)

STEP 5 ∴ △ABD ≅ △ACE — (A.A.S ≅ A.A.S)

STEP 6 Hence BD ≅ CE — (Corresponding sides of congruent △s)

The figure is a bow-tie: two straight lines crossing at C. The two triangles sit above and below the crossing point.

1 Read the figure — Two straight lines cross at C. The upper triangle has corners D, C, E; the lower one has corners A, C, B.

2 Mark what is given — Put one tick on AC and one on CE. Shade ∠B and ∠D. Notice that AC and CE are the two halves of the same straight line through C — the question is telling you C is the midpoint of AE.

3 Pick the two triangles — BC is a side of △ABC and CD is a side of △EDC. Those are the triangles to compare, written in matching order △ABC ↔ △EDC.

4 Find the free angle at C — ∠ACB and ∠ECD are vertical angles — the pair facing each other where two lines cross. Vertical angles are always equal, and you get them without being told.

5 Match three parts and finish — ∠B ≅ ∠D (given), ∠ACB ≅ ∠ECD (vertical) and AC ≅ CE (given). The side is not between the two angles, so the rule is A.A.S. ∴ △ABC ≅ △EDC, hence BC ≅ CD

GIVEN AC ≅ CE and ∠B ≅ ∠D

TO PROVE BC ≅ CD

STEP 1 In △ABC ↔ △EDC

STEP 2 ∠B ≅ ∠D — (Given)

STEP 3 ∠ACB ≅ ∠ECD — (Vertical angles)

STEP 4 AC ≅ CE — (Given)

STEP 5 ∴ △ABC ≅ △EDC — (A.A.S ≅ A.A.S)

STEP 6 Hence BC ≅ CD — (Corresponding sides of congruent △s)

Two things have to be proved here, and both come out of the same congruence.

1 Draw the isosceles triangle — AB ≅ AC, so put one tick on each of the two slanting sides. A is the vertex angle, BC the base.

2 Draw the median AD — A median runs from a corner to the midpoint of the opposite side. So mark D as the midpoint of BC: BD ≅ DC, two ticks each.

3 Split into two triangles — AD cuts the triangle into △ABD and △ACD. The side AD belongs to both — it is common.

4 Count what matches: three sides — AB ≅ AC (given), BD ≅ DC (midpoint), AD ≅ AD (common). No angle is needed — this is S.S.S. ∴ △ABD ≅ △ACD

5 Read off both results — Corresponding angles now give both answers: ∠BAD ≅ ∠CAD — so AD bisects ∠A. ∠ADB ≅ ∠ADC — and these two sit on the straight line BC, so together they make 180°. Two equal angles adding to 180° are 90° each, so AD ⊥ BC.

GIVEN In △ABC, AB ≅ AC; D is the midpoint of BC, so AD is the median from A

TO PROVE AD ⊥ BC and AD bisects ∠A

STEP 1 In △ABD ↔ △ACD

STEP 2 AB ≅ AC — (Given)

STEP 3 BD ≅ DC — (D is the midpoint of BC)

STEP 4 AD ≅ AD — (Common)

STEP 5 ∴ △ABD ≅ △ACD — (S.S.S ≅ S.S.S)

STEP 6 ∠BAD ≅ ∠CAD — (Corresponding angles of congruent △s)

STEP 7 So AD bisects ∠A

STEP 8 ∠ADB ≅ ∠ADC — (Corresponding angles of congruent △s)

STEP 9 m∠ADB + m∠ADC = 180° — (BD C is a straight line)

STEP 10 ∴ each is 90°, so AD ⊥ BC

Almost the same picture as Question 4, but this time the bisector is given and the midpoint has to be proved.

1 Draw the isosceles triangle — Same start as before: AB ≅ AC, base BC.

2 Draw the bisector of ∠A — The bisector splits ∠A into two equal parts, ∠1 ≅ ∠2. Let it meet BC at D. Do not mark BD and DC equal yet — that is the thing to prove.

3 Split into two triangles — △ABD and △ACD again share the side AD.

4 Choose S.A.S — AB ≅ AC (given), ∠1 ≅ ∠2 (bisector), AD ≅ AD (common). The equal angle sits between the two equal sides, so it is S.A.S. ∴ △ABD ≅ △ACD

5 Read off both halves of the answer — BD ≅ DC — so D is the midpoint and AD bisects the base. ∠ADB ≅ ∠ADC on a straight line — so each is 90° and AD ⊥ BC. Both together: AD is the perpendicular bisector of BC.

GIVEN In △ABC, AB ≅ AC; AD bisects ∠A, so ∠1 ≅ ∠2

TO PROVE AD is the perpendicular bisector of BC, that is BD ≅ DC and AD ⊥ BC

STEP 1 In △ABD ↔ △ACD

STEP 2 AB ≅ AC — (Given)

STEP 3 ∠1 ≅ ∠2 — (AD bisects ∠A)

STEP 4 AD ≅ AD — (Common)

STEP 5 ∴ △ABD ≅ △ACD — (S.A.S ≅ S.A.S)

STEP 6 BD ≅ DC — (Corresponding sides of congruent △s)

STEP 7 ∠ADB ≅ ∠ADC, and they make 180° — (Corresponding angles; BDC is a straight line)

STEP 8 ∴ AD ⊥ BC — (Each angle is 90°)

A square question. Everything you need comes from two facts: all four sides of a square are equal, and all four angles are 90°.

1 Draw the square — Name the corners P, Q, R, S in order around the square. Every side is equal and every corner is 90°.

2 Mark the three midpoints — X on PQ, Y on QR, Z on RS. A midpoint halves its side, and since all sides of a square are equal, every one of these halves is the same length: PX ≅ XQ ≅ QY ≅ YR ≅ RZ ≅ SZ.

3 First prove the small pair — Look at △XQY and △ZRY: XQ ≅ RZ, ∠Q ≅ ∠R (each 90°), QY ≅ YR. By S.A.S they are congruent, so XY ≅ ZY and ∠QXY ≅ ∠RZY. Each of those angles is 45°, because each triangle is right-angled with two equal legs.

4 Get the two angles at X and Z — P, X, Q lie on one straight line, so ∠PXY = 180° − 45° = 135°. In the same way ∠SZY = 135°. So ∠PXY ≅ ∠SZY.

5 Finish with S.A.S — In △PXY ↔ △SZY: PX ≅ SZ (halves of equal sides), ∠PXY ≅ ∠SZY (each 135°), XY ≅ ZY (proved in step 3). The angle lies between the two sides. ∴ △PXY ≅ △SZY

GIVEN PQRS is a square; X, Y, Z are the midpoints of PQ, QR and RS

TO PROVE △PXY ≅ △SZY

STEP 1 In △XQY ↔ △ZRY: XQ ≅ RZ, ∠Q ≅ ∠R, QY ≅ YR — (Halves of equal sides; each 90°; halves of equal sides)

STEP 2 ∴ △XQY ≅ △ZRY, so XY ≅ ZY and ∠QXY ≅ ∠RZY = 45° — (S.A.S ≅ S.A.S, then CPCTC)

STEP 3 ∠PXY = 180° − 45° = 135° = ∠SZY — (PXQ and SZR are straight lines)

STEP 4 In △PXY ↔ △SZY: PX ≅ SZ — (Halves of equal sides)

STEP 5 ∠PXY ≅ ∠SZY — (Each is 135°)

STEP 6 XY ≅ ZY — (Proved)

STEP 7 ∴ △PXY ≅ △SZY — (S.A.S ≅ S.A.S)

The only H.L question in the exercise. Two perpendicular signs in the “Given” are the clue.

1 Read the figure — A kite shape: A at the top, C at the bottom, B and D on the sides. The two perpendicular signs tell you there are right angles at B and at D.

2 Mark the right angles and the equal sides — Small squares at B and at D. One tick on AB and one on AD. Both triangles are now right-angled triangles.

3 Spot the shared hypotenuse — AC faces the right angle in both triangles, so it is the hypotenuse of each — and it is the same segment in both. That is the free match every kite question gives you.

4 Pull the two right triangles apart — △ABC in blue, △ADC in red, joined along AC.

5 Apply H.L — Hypotenuse AC common, leg AB ≅ leg AD, and each triangle has a right angle. That is exactly what H.L asks for. ∴ △ABC ≅ △ADC

GIVEN AB ⊥ BC, AD ⊥ DC and AB ≅ AD

TO PROVE △ABC ≅ △ADC

STEP 1 In △ABC ↔ △ADC

STEP 2 ∠ABC ≅ ∠ADC — (Each is a right angle: AB ⊥ BC, AD ⊥ DC)

STEP 3 AC ≅ AC — (Common hypotenuse)

STEP 4 AB ≅ AD — (Given)

STEP 5 ∴ △ABC ≅ △ADC — (H.L ≅ H.L)

The one numerical question in the exercise. No proof table is needed — just the property of a rectangle's diagonals, then one small equation.

1 Read the figure — QUAD is a rectangle with the diagonals QA and DU drawn in. They meet at C.

2 Use the first property — In a rectangle the two diagonals are equal in length: QA ≅ DU. (Each diagonal cuts the rectangle into two congruent right triangles, and the diagonals are their hypotenuses.)

3 Use the second property — The diagonals also bisect each other, so C is the midpoint of both. QC is half of QA, and DC is half of DU. Halves of equal lengths are equal: QC = DC

4 Solve the equation — x = 3x − 8; 8 = 3x − x; 8 = 2x; x = 4; Check: QC = 4 and DC = 3(4) − 8 = 4. The two halves agree, so x = 4 is right.

“Equidistant from the sides” means the perpendicular distances from O to the three sides are equal. Distance from a point to a line is always measured along the perpendicular.

1 Draw the triangle and the two bisectors — Draw △PQR. Draw the bisector of ∠Q and the bisector of ∠R and call their meeting point O.

2 Note what O gives you — O lies on the bisector of ∠Q, so ∠OQL ≅ ∠OQM. O also lies on the bisector of ∠R, so ∠ORM ≅ ∠ORN. Two pairs of equal angles, free of charge.

3 Drop the three perpendiculars — Draw OL ⊥ PQ, OM ⊥ QR and ON ⊥ RP. These three lengths are the distances we must show are equal.

4 First pair of triangles — In △OLQ ↔ △OMQ: ∠OQL ≅ ∠OQM (bisector), ∠OLQ ≅ ∠OMQ (each 90°), OQ ≅ OQ (common). Angle, angle, non-included side — A.A.S. So OL ≅ OM.

5 Second pair, and the conclusion — The same argument on the other side: in △OMR ↔ △ONR, ∠ORM ≅ ∠ORN, the right angles match and OR is common, so OM ≅ ON. Putting the two results together: OL ≅ OM ≅ ON. O is the same distance from all three sides.

6 The point O has a name: it is the incentre of the triangle, the centre of the circle that touches all three sides.

GIVEN In △PQR, the bisectors of ∠Q and ∠R meet at O

TO PROVE O is equidistant from the three sides of the triangle

CONSTRUCTION From O drop perpendiculars OL ⊥ PQ, OM ⊥ QR and ON ⊥ RP.

STEP 1 In △OLQ ↔ △OMQ: ∠OQL ≅ ∠OQM — (QO bisects ∠Q)

STEP 2 ∠OLQ ≅ ∠OMQ — (Each is a right angle)

STEP 3 OQ ≅ OQ — (Common)

STEP 4 ∴ △OLQ ≅ △OMQ, so OL ≅ OM — (A.A.S ≅ A.A.S, then CPCTC)

STEP 5 Similarly △OMR ≅ △ONR, so OM ≅ ON — (A.A.S ≅ A.A.S, then CPCTC)

STEP 6 ∴ OL ≅ OM ≅ ON — (Transitive property)

This one starts with arithmetic, not with congruent triangles. Find the angles first, then use Theorem 10.2.

1 Draw and name — Draw △ABC with A on the left and C on the right, so that AC is the base.

2 Put in the two given angles — 35° at A and 100° at B. Nothing is said about C yet.

3 Draw the perpendicular BD — From B drop a perpendicular to AC and call the foot D. This gives a right angle at D and splits △ABC into △ABD and △BDC.

4 Find ∠C — The three angles of a triangle add up to 180°: m∠C = 180° − 35° − 100° = 45°

5 Find ∠DBC — In the right triangle ABD: m∠ABD = 180° − 90° − 35° = 55°. ∠ABD and ∠DBC together make the whole angle B, so; m∠DBC = 100° − 55° = 45°

6 Apply Theorem 10.2 — In △BDC the two angles ∠DBC and ∠C are both 45°, so they are congruent. By the converse of the isosceles triangle theorem, the sides opposite them are congruent: DC ≅ DB. ∴ △BDC is isosceles

GIVEN In △ABC, m∠A = 35°, m∠B = 100° and BD ⊥ AC

TO PROVE △BDC is an isosceles triangle

STEP 1 m∠C = 180° − 35° − 100° = 45° — (Angle sum of △ABC)

STEP 2 m∠ADB = 90° — (BD ⊥ AC)

STEP 3 m∠ABD = 180° − 90° − 35° = 55° — (Angle sum of △ABD)

STEP 4 m∠DBC = 100° − 55° = 45° — (∠ABD + ∠DBC = ∠ABC)

STEP 5 ∠DBC ≅ ∠C — (Each is 45°)

STEP 6 ∴ DC ≅ DB, so △BDC is isosceles — (Theorem 10.2 — sides opposite equal angles)

The same figure as Exercise 10.1 Question 4, with the two results asked in the opposite order. The proof is identical, so this is the one to practise until you can write it from memory.

1 Draw the isosceles triangle — AB ≅ AC, base BC.

2 Mark the midpoint and join AD — D halves the base, so BD ≅ DC. Join A to D.

3 Split into two triangles — △ABD and △ACD share the side AD.

4 S.S.S — AB ≅ AC, BD ≅ DC, AD ≅ AD. Three sides. ∴ △ABD ≅ △ACD

5 Read off both results — ∠BAD ≅ ∠CAD, so AD bisects ∠A. And ∠ADB ≅ ∠ADC on the straight line BC, so each is 90° and AD ⊥ BC.

GIVEN △ABC is isosceles with AB ≅ AC; D is the mid-point of the base BC

TO PROVE AD bisects ∠A and AD ⊥ BC

STEP 1 In △ABD ↔ △ACD

STEP 2 AB ≅ AC — (Given)

STEP 3 BD ≅ DC — (D is the mid-point of BC)

STEP 4 AD ≅ AD — (Common)

STEP 5 ∴ △ABD ≅ △ACD — (S.S.S ≅ S.S.S)

STEP 6 ∠BAD ≅ ∠CAD, so AD bisects ∠A — (Corresponding angles of congruent △s)

STEP 7 ∠ADB ≅ ∠ADC = 90°, so AD ⊥ BC — (Corresponding angles; BDC is a straight line)