Mathematics • Year 8 • Unit 3 • Lesson 19
Congruence — Mixed Challenge
Pull everything from Lesson 19 together: choosing the right test (SSS, SAS, AAS, RHS), writing correct congruence statements, and using congruence to find unknowns. Six mixed problems, one "find the mistake", and one open-ended proof.
1. Mixed problems — choose the right test
Each question pulls a different idea from Lesson 19. Decide which test (SSS, SAS, AAS, RHS) applies before you start writing. Show your working and write the congruence statement where asked. 3 marks each
1.1 Triangles ABC and DEF have AB = DE = 7 cm, BC = EF = 10 cm, AC = DF = 12 cm. State the test and write the congruence statement.
1.2 In △PQR and △XYZ, PQ = XY = 9 cm, ∠Q = ∠Y = 65°, QR = YZ = 11 cm. State the test and write the congruence statement.
1.3 Two right-angled triangles each have a hypotenuse of 25 cm and one leg of 7 cm. State the test and find the length of the third side.
1.4 △KLM ≅ △PQR with ∠K = 48°, ∠L = 73°, and KM = 11 cm. State the values of ∠P, ∠Q, ∠R, and the length PR. (Hint: think about the angle sum.)
1.5 Two triangles have angles 40° and 80° and a corresponding side of 6 cm. State which test applies and explain why we don't need to know the other sides to prove congruence.
1.6 An isosceles triangle ABC has AB = AC, and M is the midpoint of BC. Prove △ABM ≅ △ACM, stating the test used and listing the three equal pairs. (Hint: the segment AM is shared by both small triangles.)
2. Find the mistake
A Year 8 student has tried to prove two triangles congruent. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — prove △ABC ≅ △DEF:
Given: AB = DE = 8 cm, AC = DF = 6 cm, ∠B = ∠E = 50°.
Line 1: We have two sides equal (AB = DE, AC = DF).
Line 2: We also have one angle equal (∠B = ∠E = 50°).
Line 3: Two sides + one angle = SAS.
Line 4: Therefore △ABC ≅ △DEF by SAS.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) State what the student actually has (which "fake test"), and what extra information would be needed to fix the proof.
Stuck? Revisit lesson § Card 5 — for SAS, the angle must be the INCLUDED angle (between the two given sides). ∠B is between AB and BC, not between AB and AC.3. Open-ended challenge — proving a parallelogram
This question has more than one valid path. 4 marks
3.1 In quadrilateral ABCD, AB = CD and AB ∥ CD. Draw the diagonal AC (it acts as a shared side for the two triangles △ABC and △CDA).
(a) Prove △ABC ≅ △CDA, listing the three equal pairs and naming the test.
(b) Use the congruence to explain why AD = BC.
(c) What type of quadrilateral must ABCD therefore be? Justify in one sentence.
(d) Bonus: find a second valid congruence test (or an alternative argument) that proves △ABC ≅ △CDA differently.
Hint: ∠BAC = ∠DCA because they are alternate angles (AB ∥ CD with transversal AC). AC is shared.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — ABC and DEF (3 sides)
All three pairs of sides equal → SSS. Vertex pairing A↔D, B↔E, C↔F, so △ABC ≅ △DEF (SSS).
1.2 — PQR and XYZ (SAS)
PQ = XY, ∠Q = ∠Y (the included angle, between PQ and QR), QR = YZ → SAS. Vertex pairing P↔X, Q↔Y, R↔Z, so △PQR ≅ △XYZ (SAS).
1.3 — Right triangles, hypotenuse 25, leg 7
Right angle + hypotenuse + one other side → RHS.
Third side² = 25² − 7² = 625 − 49 = 576, so third side = √576 = 24 cm. (7-24-25 triple.)
1.4 — △KLM ≅ △PQR
Vertex pairing K↔P, L↔Q, M↔R.
∠P = ∠K = 48°; ∠Q = ∠L = 73°.
∠R = ∠M = 180° − 48° − 73° = 59° (angle sum).
PR corresponds to KM, so PR = 11 cm.
1.5 — angles 40° and 80°, one corresponding side
Two angles plus a corresponding side → AAS. Two angles fix the SHAPE (the third angle is forced: 180° − 40° − 80° = 60°), and the one matching side fixes the SIZE — so the triangle is fully determined.
1.6 — Isosceles △ABC with midpoint M
In △ABM and △ACM:
• AB = AC (isosceles, given)
• BM = CM (M is midpoint of BC)
• AM = AM (shared / common side)
Three pairs of sides equal → △ABM ≅ △ACM (SSS).
2 — Find the mistake
(a) The mistake is on Line 3 (and is repeated in Line 4).
(b) "Two sides + one angle" alone is NOT SAS — for SAS, the angle must be the included angle (between the two given sides). The two given sides here are AB and AC, so the included angle would be ∠A. But the student was given ∠B, which is between AB and BC, not between AB and AC. So this is really an SSA configuration, not SAS.
(c) The student actually has SSA, which is NOT a valid congruence test (it allows two different triangles to share the same data). To fix the proof, the student would need either (i) ∠A = ∠D = some value (to give true SAS with sides AB and AC), or (ii) the third side BC = EF (to give SSS).
3 — Parallelogram proof (sample solution)
(a) Prove △ABC ≅ △CDA.
• AB = CD (given)
• ∠BAC = ∠DCA (alternate angles, AB ∥ CD with transversal AC)
• AC = CA (common / shared side)
Two sides plus the included angle → △ABC ≅ △CDA (SAS).
(b) Since △ABC ≅ △CDA, corresponding sides are equal: BC corresponds to DA, so AD = BC.
(c) Both pairs of opposite sides are equal (AB = CD given, AD = BC just proven), and one pair is parallel — so ABCD is a parallelogram. (In fact, opposite-sides-equal alone is enough.)
(d) Alternative approach: use AAS — ∠BAC = ∠DCA (alternate angles, proven above), ∠BCA = ∠DAC (alternate angles, BC ∥ AD once we know it), AC = CA (shared). Or: once we've proven the figure is a parallelogram, the diagonals bisect each other, giving an SAS proof on the two halves at the centre.
Marking: 1 mark for the three equal pairs correctly identified; 1 mark for the correct test name (SAS); 1 mark for (b) and (c) using corresponding sides; 1 bonus mark for a valid alternative approach in (d).