Mathematics • Year 10 • Unit 3 • Lesson 8

Congruence of Triangles — Skill Drill

Build fluency with the four congruence tests from Lesson 8: SSS, SAS, AAS, RHS. Learn to recognise which test applies, why SSA is never valid, and how to write the congruence statement with correctly matched vertices. One step at a time, from a fully worked example to independent practice.

Build · I Do / We Do / You Do

1. I do — fully worked example

Read every step. Each one has a short reason on the right so you can see why, not just what.

Problem. In triangles ABC and DEF, AB = DE = 5 cm, BC = EF = 7 cm, and ∠B = ∠E = 50°. Decide which congruence test applies and write the congruence statement.

Step 1 — List what you know.

AB = DE (a pair of sides). BC = EF (a pair of sides). ∠B = ∠E (a pair of angles).

Reason: write the matching parts before choosing a test.

Step 2 — Check where the angle sits.

∠B is between AB and BC. ∠E is between DE and EF. The angle is INCLUDED.

Reason: SAS needs the angle between the two known sides. SSA is never valid.

Step 3 — Name the test.

Two sides + included angle → SAS (Side-Angle-Side).

Step 4 — Write the statement, matching vertices.

A ↔ D, B ↔ E, C ↔ F. So △ABC ≡ △DEF (SAS).

Reason: the order of vertices in the statement is a contract — read off the equal parts in matching order.

Answer: △ABC ≡ △DEF (SAS).

Stuck? Revisit lesson § "The Four Tests" — Worked Example 1.

2. We do — fill in the missing steps

Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks

Problem. ABCD is a parallelogram with diagonal AC drawn. Prove △ABC ≡ △CDA.

Step 1 — State the two triangles being compared.

In triangles __________ and __________:

Step 2 — List the first pair of equal sides with a reason.

AB = CD (____________________ sides of a parallelogram)

Step 3 — List the second pair of equal sides with a reason.

BC = DA (____________________ sides of a parallelogram)

Step 4 — Identify the shared side.

AC = CA (____________________ side)

Step 5 — State the conclusion with the test name in brackets.

∴ △ABC ≡ △CDA (________)

Stuck? Revisit lesson § "Match the Vertices" — Worked Example 2.

3. You do — independent practice

Show your working in the space under each problem. The first four are foundation (name the test). The middle two are standard (write a short formal proof). The last two are extension (SSA trap; use CPCTC to find an unknown).

Foundation — name the test

3.1 Two triangles have sides 4 cm, 6 cm, 8 cm and 4 cm, 6 cm, 8 cm respectively. Which test proves them congruent?    1 mark

3.2 In △PQR and △XYZ, PQ = XY, ∠P = ∠X, ∠Q = ∠Y. Which test proves them congruent?    1 mark

3.3 Two right-angled triangles have hypotenuse 13 cm and one other leg 5 cm. Which test proves them congruent?    1 mark

3.4 If △ABC ≡ △DEF, list the three pairs of corresponding sides and the three pairs of corresponding angles.    1 mark

Standard — short formal proofs

3.5 In △LMN and △XYZ, LM = XY = 6 cm, MN = YZ = 9 cm, LN = XZ = 11 cm. Write a formal congruence proof.    2 marks

3.6 Two right-angled triangles, △PQR (right angle at Q) and △STU (right angle at T), have PR = SU = 17 cm (hypotenuses) and PQ = ST = 15 cm. Write a formal congruence proof.    2 marks

Extension — SSA trap and CPCTC

3.7 A student writes "In △ABC and △DEF, AB = DE, BC = EF, ∠A = ∠D, ∴ △ABC ≡ △DEF (SSA)". Explain why this is wrong, and what would have to be different for the proof to be valid.    2 marks

3.8 In a kite ABCD where AB = AD and CB = CD, the diagonal AC is drawn. (a) Prove △ABC ≡ △ADC. (b) Hence, find ∠BAC if ∠DAC = 28°.    3 marks

Stuck on 3.8? Three pairs of sides: AB = AD (given), CB = CD (given), AC = AC (common). What test does that give? Then CPCTC.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (parallelogram diagonal)

Step 1: In triangles ABC and CDA:
Step 2: AB = CD (opposite sides of a parallelogram).
Step 3: BC = DA (opposite sides of a parallelogram).
Step 4: AC = CA (common side).
Step 5: ∴ △ABC ≡ △CDA (SSS).

3.1 — Three matching sides

All three pairs of sides equal → SSS (Side-Side-Side).

3.2 — Two angles + one corresponding side

Two angles and a side → AAS (Angle-Angle-Side). The side PQ is between the two given angles, but AAS does not require the side to be included.

3.3 — Right triangle, hypotenuse + leg

Right angle + hypotenuse + one corresponding leg → RHS (Right-angle-Hypotenuse-Side).

3.4 — Corresponding parts of △ABC ≡ △DEF

Sides: AB = DE, BC = EF, AC = DF.
Angles: ∠A = ∠D, ∠B = ∠E, ∠C = ∠F.
The order of letters in the statement tells you which vertex matches which.

3.5 — Proof for △LMN ≡ △XYZ

In triangles LMN and XYZ:
LM = XY = 6 cm (given)
MN = YZ = 9 cm (given)
LN = XZ = 11 cm (given)
∴ △LMN ≡ △XYZ (SSS).

3.6 — Proof for △PQR ≡ △STU

In triangles PQR and STU:
∠Q = ∠T = 90° (given right angles)
PR = SU = 17 cm (hypotenuses)
PQ = ST = 15 cm (corresponding legs)
∴ △PQR ≡ △STU (RHS).

3.7 — Why "SSA" is wrong

SSA is not a valid congruence test. The angle ∠A is not between the two known sides AB and BC — it sits at vertex A, while one of the sides (BC) is on the opposite side of the triangle. This is the "ambiguous case" because two different triangles can share the same two sides and a non-included angle.
For the proof to be valid, either (i) the equal angle would need to be ∠B (between AB and BC), giving SAS, or (ii) the two triangles would need to be right-angled at the unmentioned vertex with one of the sides being the hypotenuse, giving RHS.

3.8 — Kite proof and ∠BAC

(a) In triangles ABC and ADC:
AB = AD (given)
CB = CD (given)
AC = AC (common side)
∴ △ABC ≡ △ADC (SSS).
(b) By CPCTC (corresponding parts of congruent triangles), ∠BAC = ∠DAC = 28°.
This shows the diagonal AC of a kite bisects the apex angle ∠BAD.