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.