Mathematics • Year 10 • Unit 3 • Lesson 10
Proving Congruence — Skill Drill
Build fluency with the Lesson 10 proof structure: name the two triangles → list each matching part with a reason → state the test in brackets → finish with CPCTC if asked for a follow-up. One step at a time, from a fully worked example to independent practice with all four valid tests (SSS, SAS, AAS, RHS).
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. Prove △ABC ≡ △DEF, given AB = DE, BC = EF, ∠B = ∠E.
Step 1 — Name the triangles.
In triangles ABC and DEF:
Reason: examiners want every proof to start by naming both triangles in the same vertex order as the conclusion.
Step 2 — List the matching parts with reasons.
AB = DE (given)
∠B = ∠E (given)
BC = EF (given)
Reason: write each pair on a new line, with the reason in brackets. The order matters because we want the side-angle-side pattern visible.
Step 3 — Verify the angle is INCLUDED.
∠B is between sides AB and BC. ∠E is between sides DE and EF.
Reason: SAS requires the angle to be between the two known sides. Without checking this, we risk SSA — which is not valid.
Step 4 — Write the conclusion with the test in brackets.
∴ △ABC ≡ △DEF (SAS)
Reason: the symbol ∴ means "therefore". The test name in brackets is required for full marks.
Answer: △ABC ≡ △DEF (SAS).
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 and hence find ∠BAC if ∠DCA = 35°.
Step 1 — Name the two triangles.
In triangles ____________ and ____________:
Step 2 — List the matching sides with reasons.
AB = CD (____________________ sides of parallelogram)
BC = DA (____________________ sides of parallelogram)
AC = CA (____________________ side)
Step 3 — State the congruence with the test in brackets.
∴ △ABC ≡ △CDA (________)
Step 4 — Use CPCTC to find ∠BAC.
∠BAC = ∠________ (corresponding angles of congruent triangles)
∴ ∠BAC = ______°
3. You do — independent practice
Write a full formal proof for each. Every matching part needs a reason in brackets; every conclusion needs the test name in brackets. The first four are foundation (one valid test). The middle two are standard (proof + one CPCTC step). The last two are extension (the ambiguous case; a proof requiring two reasons per line).
Foundation — name the test and write the proof
3.1 In △ABC and △DEF, AB = DE, BC = EF, AC = DF. Write the proof. 1 mark
3.2 In △PQR and △XYZ, ∠P = ∠X, ∠R = ∠Z, PQ = XY. Write the proof. 1 mark
3.3 In △ABC (right-angled at B) and △DEF (right-angled at E), AC = DF = 17 cm (hypotenuses) and AB = DE = 8 cm. Write the proof. 1 mark
3.4 In △KLM and △NPQ, KL = NP, LM = PQ, ∠L = ∠P. Write the proof. 1 mark
Standard — proof plus CPCTC
3.5 In △ABC, AB = AC and AD is the perpendicular from A to BC (with D on BC). (a) Prove △ABD ≡ △ACD. (b) Hence prove BD = DC. 2 marks
3.6 Line segments AB and CD bisect each other at point M (so AM = MB and CM = MD). (a) Prove △AMC ≡ △BMD. (b) Hence deduce AC = BD. 2 marks
Extension — the ambiguous case and a richer proof
3.7 A student writes: "In △ABC and △DEF, AB = DE = 7, BC = EF = 10, ∠A = ∠D = 35°, ∴ △ABC ≡ △DEF (SAS)". Explain in 1–2 sentences why this is wrong, and state the correct conclusion you can reach about these two triangles. 2 marks
3.8 ABCD is a rectangle. (a) Prove △ABD ≡ △BAC. (b) Hence prove that the diagonals AC and BD are equal in length. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (parallelogram + CPCTC)
Step 1: In triangles ABC and CDA:
Step 2: AB = CD (opposite sides), BC = DA (opposite sides), AC = CA (common).
Step 3: ∴ △ABC ≡ △CDA (SSS).
Step 4: ∠BAC = ∠DCA (corresponding angles of congruent triangles) → ∠BAC = 35°.
3.1 — Three matching sides
In triangles ABC and DEF:
AB = DE (given), BC = EF (given), AC = DF (given).
∴ △ABC ≡ △DEF (SSS).
3.2 — AAS
In triangles PQR and XYZ:
∠P = ∠X (given), ∠R = ∠Z (given), PQ = XY (given corresponding side).
∴ △PQR ≡ △XYZ (AAS).
3.3 — RHS
In triangles ABC and DEF:
∠B = ∠E = 90° (given), AC = DF = 17 cm (hypotenuses), AB = DE = 8 cm (corresponding legs).
∴ △ABC ≡ △DEF (RHS).
3.4 — SAS (included angle)
In triangles KLM and NPQ:
KL = NP (given), ∠L = ∠P (given — between KL and LM, and between NP and PQ), LM = PQ (given).
∴ △KLM ≡ △NPQ (SAS).
3.5 — Isosceles with perpendicular from apex
(a) In triangles ABD and ACD:
AB = AC (given)
∠ADB = ∠ADC = 90° (AD is the perpendicular to BC)
AD = AD (common side)
∴ △ABD ≡ △ACD (RHS).
(b) ∴ BD = DC (CPCTC). The perpendicular from the apex of an isosceles triangle bisects the base.
3.6 — Segments bisecting each other
(a) In triangles AMC and BMD:
AM = BM (given — AB bisected at M)
∠AMC = ∠BMD (vertically opposite angles)
CM = DM (given — CD bisected at M)
∴ △AMC ≡ △BMD (SAS).
(b) ∴ AC = BD (CPCTC).
3.7 — Spotting the SSA trap
The student's labelling is SSA, not SAS. The angle ∠A sits at vertex A, but the two given sides AB and BC meet at vertex B, not A — so the angle is not included between the two known sides. This is the ambiguous case and two non-congruent triangles can be drawn from the same measurements. The correct conclusion is that no valid congruence test applies with this combination of givens.
3.8 — Rectangle diagonals equal
(a) In triangles ABD and BAC:
AB = BA (common side)
AD = BC (opposite sides of rectangle)
∠DAB = ∠CBA = 90° (corners of rectangle)
∴ △ABD ≡ △BAC (SAS).
(b) ∴ BD = AC (CPCTC). The two diagonals of a rectangle are equal in length.