Mathematics • Year 7 • Unit 3 • Lesson 15
Applying Congruence — Mixed Challenge
Pull together everything from Lessons 14 and 15: prove congruence, deduce equal sides/angles, isosceles base angles, parallelogram diagonal, plus the classic Year-7 reverse mistake (claiming a result without proving congruence first).
1. Mixed problems
Show working. Cite test names AND the "(matching … of ≡ △s)" reason for deduced parts. 2 marks each
1.1 △ABC ≡ △XYZ with AB = 12 cm, BC = 9 cm, ∠B = 65°. Find XY, YZ and ∠Y.
1.2 △ABC has AB = AC = 10 cm. M is the midpoint of BC. Without measuring, state ∠ABM ÷ ∠ACM (give the ratio). Briefly justify by congruence.
1.3 ABCD is a parallelogram. Diagonal BD is drawn. (a) Which test proves △ABD ≡ △CDB? (b) Deduce one pair of equal sides.
1.4 △PQR ≡ △STU with PQ = 4x + 2 cm and ST = 2x + 14 cm. Find x and PQ.
1.5 △ABC ≡ △DEF. ∠A = (3y + 10)° and ∠D = (5y − 20)°. Find y and the value of ∠A.
1.6 Lines AB and CD intersect at point O. △AOC and △BOD are formed. Given that OA = OB and OC = OD (i.e. O is the midpoint of both AB and CD). (a) Which test proves △AOC ≡ △BOD? (b) Deduce AC = BD with reason.
2. Find the mistake
Exactly one step contains a mistake. Spot it, explain, then redo. 3 marks
Student's question: △ABC has AB = AC. Prove ∠B = ∠C.
Step 1: Given: AB = AC.
Step 2: Therefore ∠B = ∠C (matching ∠s of ≡ △s).
Step 3: ∴ ∠B = ∠C.
(a) What's wrong with this proof?
(b) Explain why the conclusion is not yet justified.
(c) Write the full proof with all required steps.
Stuck? Before you can write "matching ∠s of ≡ △s", you have to PROVE two triangles are congruent. The student skipped that step entirely — there's no second triangle in sight!3. Open-ended challenge — prove diagonals of a rhombus bisect each other
Multiple valid proof routes exist. Show your reasoning clearly. 4 marks
3.1 A rhombus ABCD has all four sides equal: AB = BC = CD = DA. The two diagonals AC and BD intersect at point O.
Prove that the diagonals bisect each other — i.e. show that OA = OC AND OB = OD.
Steps to follow:
(i) Identify two sub-triangles to prove congruent (e.g. △AOB and △COD, or △AOB and △COB).
(ii) List the equal parts and cite reasons (sides equal because rhombus, alternate angles because opposite sides are parallel, common side, etc.).
(iii) Name the test (SSS/SAS/AAS/RHS).
(iv) Write the congruence statement.
(v) Deduce OA = OC and OB = OD with the reason "(matching sides of ≡ △s)".
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Read off matching parts
XY = AB = 12 cm, YZ = BC = 9 cm, ∠Y = ∠B = 65° (matching sides/∠s of ≡ △s).
1.2 — Ratio ∠ABM ÷ ∠ACM
In △ABM and △ACM: AB = AC (given), BM = CM (M midpoint), AM = AM (common) → △ABM ≡ △ACM (SSS). So ∠ABM = ∠ACM (matching ∠s of ≡ △s). Ratio = 1 : 1 (they are equal).
1.3 — Parallelogram diagonal BD
(a) In △ABD and △CDB: ∠ABD = ∠CDB (alt. ∠s, AB ∥ DC), ∠ADB = ∠CBD (alt. ∠s, AD ∥ BC), BD = DB (common) → △ABD ≡ △CDB (AAS).
(b) ∴ AB = CD (matching sides of ≡ △s). (Equally valid: AD = CB.)
1.4 — △PQR ≡ △STU algebra
Matching sides: 4x + 2 = 2x + 14 (matching sides of ≡ △s).
2x = 12 → x = 6. PQ = 4(6) + 2 = 26 cm. Check ST = 2(6) + 14 = 26 ✓.
1.5 — △ABC ≡ △DEF angle algebra
Matching angles: 3y + 10 = 5y − 20 (matching ∠s of ≡ △s).
30 = 2y → y = 15. ∠A = 3(15) + 10 = 55°. Check ∠D = 5(15) − 20 = 55° ✓.
1.6 — Lines intersecting at O
(a) In △AOC and △BOD: OA = OB (given), ∠AOC = ∠BOD (vert. opp. ∠s), OC = OD (given) → two sides + included angle → SAS. ∴ △AOC ≡ △BOD (SAS).
(b) AC = BD (matching sides of ≡ △s).
2 — Find the mistake
(a) The student went STRAIGHT from "AB = AC" to "∠B = ∠C" without proving two triangles congruent first. The reason "(matching ∠s of ≡ △s)" is meaningless because no two triangles have been shown to be congruent — there isn't even a second triangle to refer to.
(b) The "(matching ∠s of ≡ △s)" reason can ONLY be cited AFTER you have proved △X ≡ △Y by one of the four tests. The student jumped to the conclusion.
(c) Corrected proof:
Step 1: Let M be the midpoint of BC. Draw AM.
Step 2: In △ABM and △ACM:
AB = AC (given)
BM = CM (M midpoint of BC)
AM = AM (common)
Step 3: △ABM ≡ △ACM (SSS).
Step 4: ∴ ∠B = ∠C (matching ∠s of ≡ △s). ✓
3 — Rhombus diagonals bisect each other (sample proof)
In △AOB and △COD:
AB = CD (sides of rhombus equal)
∠OAB = ∠OCD (alt. ∠s, AB ∥ DC since rhombus is also a parallelogram)
∠OBA = ∠ODC (alt. ∠s, AB ∥ DC)
Two angles + a matching side → △AOB ≡ △COD (AAS).
∴ OA = OC (matching sides of ≡ △s)
∴ OB = OD (matching sides of ≡ △s)
Therefore the diagonals AC and BD bisect each other at O. ✓
Alternative proof using SSS: In △AOB and △COB (sharing side OB): AB = CB (rhombus sides equal), OB = OB (common). To get the third side, use the diagonal property of a parallelogram, OR show △AOB ≡ △COB via the diagonals chase. There are several valid routes — accept any rigorous one.
Marking: 1 for identifying valid sub-triangles; 1 for listing equal parts with reasons; 1 for naming the test and congruence statement; 1 for deducing OA = OC and OB = OD with correct reason.