Mathematics • Year 10 • Unit 3 • Lesson 10
Proving Congruence — Mixed Challenge
Pull together every idea from Lesson 10: the four valid tests (SSS, SAS, AAS, RHS), the SSA ambiguous case, full formal proof structure with reasons, vertex matching, and CPCTC. Spot a classmate's plausible mistake, then design your own multi-step proof problem.
1. Mixed problems — write a full proof
Each question uses a different idea from Lesson 10. Reasons in brackets after every matching part; test in brackets after the conclusion. 3 marks each
1.1 In △ABC and △DEF, AB = DE, ∠A = ∠D, AC = DF. Write a full proof and name the test.
1.2 In △PQR and △XYZ, ∠P = ∠X = 90°, PQ = XY (one pair of legs), and QR = YZ (the hypotenuses). Write a full proof and name the test.
1.3 If △ABC ≡ △DEF (proved by AAS) and AB = 6 cm, BC = 9 cm, ∠B = 70°, ∠A = 50°, list every known side and angle of △DEF, with a brief reason for the angle not directly given.
1.4 Two segments AB and CD intersect at M, with AM = BM and CM = DM. Prove △AMC ≡ △BMD and hence AC = BD.
1.5 In an isosceles triangle PQR with PQ = PR, the bisector of ∠P meets QR at M. Prove △PQM ≡ △PRM and hence prove PM ⊥ QR.
1.6 ABCD is a square with diagonals AC and BD meeting at M. Prove △AMB ≡ △CMB and hence prove that the diagonals are perpendicular (∠AMB = 90°).
2. Find the mistake
Another Year 10 student is asked: "Given AB = DE, BC = EF, ∠A = ∠D, prove △ABC ≡ △DEF." Their working is below. Exactly one step contains a mistake. Spot it, explain why it's wrong, then write a correct response. 3 marks
Student's working:
Line 1: In triangles ABC and DEF:
Line 2: AB = DE (given)
Line 3: ∠A = ∠D (given — between AB and BC)
Line 4: BC = EF (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 showing why no valid proof is possible with these givens, or stating what would need to change for the proof to work.
Stuck? ∠A sits at vertex A. AB and BC meet at vertex B, not at A. So ∠A is NOT between AB and BC — it's between AB and AC. The student wrote a wrong description of where the angle sits.3. Open-ended challenge — design a two-stage congruence proof
This question has many valid answers. Be creative but show every reason. 4 marks
3.1 Design a two-stage congruence problem set in a real or geometric context. Your problem must:
(i) Involve a familiar shape (parallelogram, rectangle, rhombus, kite, square, isosceles triangle, or two intersecting segments).
(ii) Ask the solver to prove congruence of two triangles using one of the four valid tests, then use CPCTC to deduce a specific property (equal sides, equal angles, perpendicularity, bisection, etc.).
(iii) Include a labelled rough diagram so the reader can see which vertices match.
Then write the full formal proof of your own problem with every reason in brackets, ending with the CPCTC step. Conclude with a one-sentence statement of what your proof has established.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Two sides with included angle
In triangles ABC and DEF:
AB = DE (given)
∠A = ∠D (given — between AB and AC)
AC = DF (given)
∴ △ABC ≡ △DEF (SAS).
∠A sits at vertex A and is between AB and AC, so it IS the included angle here — not between AB and BC.
1.2 — Right triangle + hypotenuse + leg
In triangles PQR and XYZ:
∠P = ∠X = 90° (given)
QR = YZ (hypotenuses)
PQ = XY (corresponding legs)
∴ △PQR ≡ △XYZ (RHS).
1.3 — Listing every part of △DEF
By CPCTC: DE = AB = 6 cm, EF = BC = 9 cm, ∠D = ∠A = 50°, ∠E = ∠B = 70°.
Third angle: ∠F = ∠C = 180° − 50° − 70° = 60° (angle sum of a triangle is 180°). Side DF = AC, which is not given numerically.
1.4 — Intersecting segments bisecting each other
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).
∴ AC = BD (CPCTC).
1.5 — Angle bisector from apex of isosceles triangle
In triangles PQM and PRM:
PQ = PR (given)
∠QPM = ∠RPM (given — PM bisects ∠P)
PM = PM (common side)
∴ △PQM ≡ △PRM (SAS).
By CPCTC, ∠PMQ = ∠PMR. But ∠PMQ + ∠PMR = 180° (angles on the straight line QR). So 2∠PMQ = 180°, giving ∠PMQ = 90°. ∴ PM ⊥ QR.
1.6 — Square diagonals are perpendicular
In triangles AMB and CMB:
AB = CB (all sides of square equal)
AM = CM (diagonals of square bisect each other)
BM = BM (common side)
∴ △AMB ≡ △CMB (SSS).
By CPCTC, ∠AMB = ∠CMB. But ∠AMB + ∠CMB = 180° (angles on the straight line AC). So 2∠AMB = 180°, giving ∠AMB = 90°. ∴ The diagonals of a square are perpendicular.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The student claims "∠A is between AB and BC", but this is geometrically false. Sides AB and BC meet at vertex B, not at vertex A. The angle at A (i.e. ∠A) sits between AB and AC — so it is not the included angle for the side pair (AB, BC).
(c) With the corrected reading of Line 3, the proof falls apart: we have two sides (AB and BC) and a non-included angle (∠A). This is the SSA configuration, which is not a valid congruence test. The correct response is to say no valid SAS proof can be written from these givens. To rescue the proof, the equal angle should be ∠B (the included angle between AB and BC), giving a valid SAS proof.
This is exactly the trap the Lesson 10 "ambiguous case" card warns about.
3 — Open-ended challenge (sample solution)
Problem. ABCD is a rhombus (all four sides equal). The diagonals AC and BD meet at M. Prove △ABM ≡ △ADM, and hence prove that the diagonals are perpendicular.
Diagram: rhombus with vertices A (top), B (right), C (bottom), D (left). Diagonals AC (vertical) and BD (horizontal) cross at M.
Proof.
In triangles ABM and ADM:
AB = AD (all sides of rhombus equal)
BM = DM (diagonals of a rhombus bisect each other)
AM = AM (common side)
∴ △ABM ≡ △ADM (SSS).
By CPCTC, ∠AMB = ∠AMD. But ∠AMB + ∠AMD = 180° (angles on the straight line BD), so 2∠AMB = 180°, giving ∠AMB = 90°. ∴ The diagonals of a rhombus are perpendicular.
Marking: 1 for a clear shape/context with diagram, 1 for a valid use of one of the four tests, 1 for every reason in brackets and the test in brackets, 1 for a CPCTC step that yields a specific property. Full marks for any valid two-stage proof meeting all four.