Mathematics • Year 7 • Unit 3 • Lesson 15

Applying Congruence

Once two triangles are proved congruent, EVERY pair of corresponding sides and angles is automatically equal. Use this to read off missing lengths and angles, and to prove sides or angles equal in figures like isosceles triangles and parallelograms.

Build · I Do / We Do / You Do

1. I do — fully worked example

The classic "isosceles base angles are equal" proof — once you can do this one, you can do the rest.

Problem. △ABC has AB = AC (isosceles). M is the midpoint of BC. Prove ∠B = ∠C.

Step 1 — Identify the two sub-triangles.

Drawing AM splits △ABC into △ABM and △ACM.

Step 2 — List the equal parts in △ABM and △ACM.

AB = AC (given, isosceles)

BM = CM (M is midpoint of BC)

AM = AM (common side — shared between both triangles)

Step 3 — Identify the test.

Three pairs of sides equal → SSS.

Step 4 — Write the congruence statement.

△ABM ≡ △ACM (SSS)

Step 5 — Deduce the equal angles.

∴ ∠B = ∠C (matching ∠s of ≡ △s)

Reason: once the triangles are congruent, ALL matching parts are equal — including ∠B and ∠C.

Answer: ∠B = ∠C, proved via △ABM ≡ △ACM (SSS).

Stuck? Revisit lesson § "Proving Two Angles Are Equal" — the trick is always: prove the two sub-triangles congruent first, THEN read off the matching angle.

2. We do — fill in the missing steps

Read-off-the-matching-parts question. Fill in the blanks. 4 marks

Problem. △ABC ≡ △DEF. AB = 6.5 cm, BC = 8 cm, and ∠B = 50°. Find DE, EF and ∠E.

Step 1 — Match the vertices: A ↔ ____, B ↔ ____, C ↔ ____.

Step 2 — Read off matching sides and angle:

DE matches AB, so DE = ____ cm.

EF matches BC, so EF = ____ cm.

∠E matches ∠B, so ∠E = ____°.

Step 3 — Cite the reason for each: all three follow from "(matching ____________ of ≡ △s)".

Step 4 — State the final answers: DE = ____ cm, EF = ____ cm, ∠E = ____°.

Stuck? Revisit lesson § "Watch Me Solve It · Find the missing length" — the order of vertices in the congruence statement tells you which parts match.

3. You do — independent practice

Show working. Cite reasons in brackets — "(matching sides/∠s of ≡ △s)" for the deduced parts.

Foundation — read off matching parts

3.1 △ABC ≡ △PQR with AB = 5 cm, BC = 7 cm, AC = 9 cm. State PQ, QR and PR. 1 mark

3.2 △XYZ ≡ △LMN with ∠X = 35°, ∠Y = 80°. State ∠L, ∠M and ∠N. 1 mark

3.3 △DEF ≡ △STU with DE = 4 cm, EF = 6 cm, DF = 7 cm, ∠D = 80°, ∠E = 60°. State ST, TU, SU, ∠S, ∠T, ∠U. 1 mark

3.4 △ABC ≡ △XYZ. ∠A = 40°, ∠B = 65°. Find ∠Z. 1 mark

Standard — prove congruent first, then deduce

3.5 △ABC has AB = AC. M is the midpoint of BC. Prove that AM is perpendicular to BC (i.e. ∠AMB = 90°). Use congruent triangles in your proof. 2 marks

3.6 ABCD is a parallelogram (AB ∥ DC and AD ∥ BC). Diagonal AC is drawn. Prove that △ABC ≡ △CDA. 2 marks

Extension — push your reasoning

3.7 Using the congruence from 3.6, deduce that the opposite sides of a parallelogram are equal: AB = DC and BC = AD. Cite the reason. 2 marks

3.8 △PQR ≡ △STU with PQ = 3x − 4 cm and ST = x + 8 cm. Find x and state PQ. 2 marks

Stuck on 3.8? Matching sides of congruent triangles are equal: 3x − 4 = x + 8.

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Section 2 — We do (△ABC ≡ △DEF)

Step 1: A ↔ D, B ↔ E, C ↔ F.
Step 2: DE = 6.5 cm, EF = 8 cm, ∠E = 50°.
Step 3: (matching sides/∠s of ≡ △s).
Step 4: DE = 6.5 cm, EF = 8 cm, ∠E = 50°.

3.1 — △ABC ≡ △PQR sides

PQ = AB = 5 cm; QR = BC = 7 cm; PR = AC = 9 cm (matching sides of ≡ △s).

3.2 — △XYZ ≡ △LMN angles

∠L = ∠X = 35°; ∠M = ∠Y = 80°; ∠N = 180 − 35 − 80 = 65° (angles in a triangle sum to 180°); also ∠N = ∠Z = 65° (matching ∠s of ≡ △s).

3.3 — △DEF ≡ △STU all parts

ST = DE = 4 cm; TU = EF = 6 cm; SU = DF = 7 cm; ∠S = ∠D = 80°; ∠T = ∠E = 60°; ∠U = 180 − 80 − 60 = 40° (and ∠U = ∠F) (matching sides/∠s of ≡ △s).

3.4 — △ABC ≡ △XYZ find ∠Z

∠Z matches ∠C. ∠C = 180 − 40 − 65 = 75° (∠s in a △), so ∠Z = 75° (matching ∠s of ≡ △s).

3.5 — AM ⊥ BC

In △ABM and △ACM: AB = AC (given), BM = CM (M midpoint), AM = AM (common). Three sides equal → △ABM ≡ △ACM (SSS).
∴ ∠AMB = ∠AMC (matching ∠s of ≡ △s). But ∠AMB + ∠AMC = 180° (∠s on a straight line BMC). So 2∠AMB = 180°, giving ∠AMB = 90°. ∴ AM ⊥ BC.

3.6 — Parallelogram diagonal proof

In △ABC and △CDA:
∠BAC = ∠DCA (alt. ∠s, AB ∥ DC)
∠BCA = ∠DAC (alt. ∠s, AD ∥ BC)
AC = CA (common)
Two angles + a common side → △ABC ≡ △CDA (AAS).

3.7 — Opposite sides of a parallelogram equal

From △ABC ≡ △CDA (proved in 3.6):
AB = CD (matching sides of ≡ △s).
BC = DA (matching sides of ≡ △s).
∴ opposite sides of a parallelogram are equal.

3.8 — △PQR ≡ △STU with algebra

Matching sides equal: 3x − 4 = x + 8 (matching sides of ≡ △s).
2x = 12 → x = 6.
PQ = 3(6) − 4 = 14 cm. Check: ST = 6 + 8 = 14 cm ✓.