Mathematics • Year 10 • Unit 3 • Lesson 8
Congruence of Triangles — Mixed Challenge
Pull together every idea from Lesson 8: the four valid tests (SSS, SAS, AAS, RHS); why SSA is never valid; how to match vertices in a congruence statement; and how to use CPCTC. Spot a classmate's plausible mistake, then design your own congruence problem.
1. Mixed problems — pick the right test
Each question uses a different idea from Lesson 8. State the test before writing the conclusion. Show your reasoning. 3 marks each
1.1 In △ABC and △PQR: AB = PQ, AC = PR, ∠A = ∠P. Name the test and write the congruence statement.
1.2 In △XYZ and △LMN: ∠X = ∠L, ∠Y = ∠M, XZ = LN. Name the test and write the congruence statement.
1.3 Two right triangles have right angles at B and Q respectively, hypotenuses AC = PR = 25 cm, and legs AB = PQ = 7 cm. Name the test and write the congruence statement.
1.4 If △ABC ≡ △DEF and AB = 6 cm, BC = 8 cm, ∠A = 40°, ∠C = 60°, state DE, EF, ∠D, ∠F, and ∠B and ∠E (justify the last two).
1.5 ABCD is a square. Diagonal AC is drawn. Prove △ABC ≡ △ADC and state which test you used.
1.6 In the diagram, AB and CD are two line segments that intersect at point M, with AM = DM and BM = CM. Prove that △AMB ≡ △DMC and name the test.
2. Find the mistake
Another Year 10 student is asked to prove △ABC ≡ △DEF given AB = DE, BC = EF, and ∠A = ∠D. Their working is below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then write the corrected response. 3 marks
Student's working:
Line 1: In triangles ABC and DEF:
Line 2: AB = DE (given)
Line 3: BC = EF (given)
Line 4: ∠A = ∠D (given)
Line 5: ∴ △ABC ≡ △DEF (SAS)
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write a correct response — either explain why the proof cannot be completed with these givens, or describe what would need to change. State the correct test name if a valid proof is possible.
Stuck? Where does ∠A sit? Is it between AB and BC, or somewhere else? Recall: SAS requires the angle to be INCLUDED between the two known sides.3. Open-ended challenge — design four congruence pairs
This question has many valid answers. Be creative but show every measurement. 4 marks
3.1 Design four different pairs of triangles, one for each of the four valid tests (SSS, SAS, AAS, RHS). For each pair you must:
(i) Give specific side lengths in cm and angles in degrees.
(ii) Name the two triangles using consistent vertex matching (e.g. △ABC and △DEF with A ↔ D).
(iii) Write a one-line congruence statement with the test in brackets.
(iv) For your SAS pair, clearly mark which angle is the INCLUDED angle. For your RHS pair, clearly mark the right angle and hypotenuse.
Bonus: also design one "trap" pair where SSA looks tempting but is not valid, and explain why it would fail.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — AB = PQ, AC = PR, ∠A = ∠P
∠A is between sides AB and AC (the included angle).
Test: SAS. Statement: △ABC ≡ △PQR (SAS).
1.2 — Two angles + non-included side
XZ is opposite ∠Y, not between ∠X and ∠Y. Two angles plus any corresponding side is the AAS test.
Statement: △XYZ ≡ △LMN (AAS).
1.3 — Right triangle, hypotenuse + leg
Test: RHS. Statement: △ABC ≡ △PQR (RHS).
By Pythagoras the third leg is also equal (24 cm), so this is essentially SSS in disguise — but RHS is the proper name.
1.4 — CPCTC values
Sides: DE = AB = 6 cm, EF = BC = 8 cm.
Angles: ∠D = ∠A = 40°, ∠F = ∠C = 60°.
∠B and ∠E: 180° − 40° − 60° = 80° each (angle sum of a triangle).
1.5 — Square diagonal
In triangles ABC and ADC:
AB = AD (all sides of square equal)
BC = DC (all sides of square equal)
AC = AC (common side)
∴ △ABC ≡ △ADC (SSS).
Alternative valid proof uses SAS with the right angles at B and D: AB = AD, ∠B = ∠D = 90°, BC = DC.
1.6 — Intersecting segments
In triangles AMB and DMC:
AM = DM (given)
∠AMB = ∠DMC (vertically opposite angles)
BM = CM (given)
∴ △AMB ≡ △DMC (SAS).
2 — Find the mistake
(a) The mistake is on Line 5.
(b) The student labelled the test as SAS, but the angle ∠A is not the included angle between sides AB and BC. AB and BC meet at vertex B, not vertex A. So this is the SSA configuration, which is not a valid congruence test.
(c) Correct response: the proof as written cannot be completed because two sides plus a non-included angle (SSA) is the ambiguous case. To rescue it, either (i) the given angle should be ∠B (between AB and BC), giving valid SAS, or (ii) we would need the triangles to be right-angled with one of the sides being the hypotenuse, giving RHS.
Trap: the student picked plausible-looking givens that mention the right letters but in the wrong configuration. Always check whether the angle sits between the two sides.
3 — Open-ended challenge (sample solutions)
SSS pair. △ABC with sides 5, 7, 9 cm and △DEF with sides 5, 7, 9 cm. AB = DE = 5, BC = EF = 7, CA = FD = 9. △ABC ≡ △DEF (SSS).
SAS pair. △PQR with PQ = 6, QR = 8, included angle ∠Q = 70°; △XYZ with XY = 6, YZ = 8, ∠Y = 70°. △PQR ≡ △XYZ (SAS). Included angle is ∠Q (between PQ and QR), matching ∠Y (between XY and YZ).
AAS pair. △LMN with ∠L = 40°, ∠M = 60°, side LM = 10 cm; △STU with ∠S = 40°, ∠T = 60°, side ST = 10 cm. △LMN ≡ △STU (AAS).
RHS pair. △ABC right-angled at B with hypotenuse AC = 13 cm, leg AB = 5 cm; △DEF right-angled at E with hypotenuse DF = 13 cm, leg DE = 5 cm. △ABC ≡ △DEF (RHS). The 5-12-13 triple makes the third side automatic.
Bonus — SSA trap. △GHI with GH = 6, HI = 9, ∠G = 35°; △JKL with JK = 6, KL = 9, ∠J = 35°. This looks like two sides and an angle, but ∠G is not between GH and HI — it sits at vertex G, opposite side HI. So this is SSA, the ambiguous case, and two non-congruent triangles can be drawn from these givens.
Marking: 1 mark per valid test pair (4 total). Bonus credit for the SSA trap with a clear explanation of why it fails.