Mathematics • Year 10 • Unit 3 • Lesson 10
Writing Proofs Examiners Love
Apply Lesson 10's formal proof structure to richer geometric contexts: isosceles triangles, kites, parallelograms, intersecting chords, and shape-property deductions. Every proof must (1) name the triangles, (2) list matching parts with reasons, (3) state the test, (4) use CPCTC for any follow-up. Missing a reason loses half a mark.
1. Word problems
For each problem: write a full formal proof. Reasons in brackets after every matching part; test in brackets after the congruence statement. Then complete any CPCTC step asked for.
1.1 — Isosceles roof beam. A timber A-frame has rafters AB and AC of equal length meeting at apex A. A horizontal tie-beam BC closes the base. Point M is the midpoint of BC, and a vertical post AM is added.
(a) Prove △ABM ≡ △ACM.
(b) Hence prove ∠ABC = ∠ACB (the base angles of an isosceles triangle are equal). 3 marks
1.2 — Kite hinge. A diamond-shaped kite ABCD has AB = AD and CB = CD (typical kite property). Diagonal BD is drawn.
(a) Prove △ABD is isosceles (using the given AB = AD) — state the isosceles property used.
(b) Prove △ABC ≡ △ADC using a SSS proof, and hence deduce that diagonal AC bisects ∠BAD. 3 marks
1.3 — Parallelogram diagonals bisect each other. ABCD is a parallelogram with diagonals AC and BD meeting at M.
(a) Prove △AMB ≡ △CMD.
(b) Hence prove AM = MC and BM = MD (so the diagonals of a parallelogram bisect each other). 3 marks
1.4 — Surveyor's check. Two pegs P and Q are on opposite sides of a river. To find PQ, the surveyor measures PR and angles at P and R from a third peg R on the same side as P. The same is done on a copy site with pegs P′, Q′ and R′. If PR = P′R′, ∠P = ∠P′ and ∠R = ∠R′, prove △PQR ≡ △P′Q′R′, and hence conclude PQ = P′Q′.
3 marks
1.5 — Right-angled gables. Two right-angled triangular gables on adjacent houses both have a vertical post 3 m tall and a sloping rafter (the hypotenuse) 5 m long, with the right angle at the foot of the post.
(a) Prove the two gables are congruent.
(b) Hence deduce that the horizontal base of each gable is the same length, without using Pythagoras explicitly — just CPCTC. 3 marks
2. Explain your thinking
This question is about reasoning, not just numbers. Use full sentences. 4 marks
2.1 A friend says: "I always just write '△ABC ≡ △DEF' at the end and skip the part where I list the matching pieces and the test — the diagram makes it obvious." Using Lesson 10, explain (i) what is reasonable about the friend's instinct, (ii) why this loses marks even when the conclusion is right, (iii) the four required parts of a formal proof (in the order they should appear), and (iv) why CPCTC is sometimes essential after the conclusion. Use the phrase "test in brackets" somewhere in your answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Isosceles roof beam
(a) In triangles ABM and ACM:
AB = AC (given — equal rafters)
BM = CM (M is midpoint of BC)
AM = AM (common side)
∴ △ABM ≡ △ACM (SSS).
(b) By CPCTC, ∠ABM = ∠ACM, i.e. ∠ABC = ∠ACB. This proves the base angles of an isosceles triangle are equal.
1.2 — Kite diagonal bisects apex angle
(a) Since AB = AD (given), triangle ABD has two equal sides, which is the definition of isosceles. ∴ △ABD is isosceles.
(b) In triangles ABC and ADC:
AB = AD (given)
CB = CD (given)
AC = AC (common side)
∴ △ABC ≡ △ADC (SSS).
By CPCTC, ∠BAC = ∠DAC. So diagonal AC bisects ∠BAD.
1.3 — Parallelogram diagonals bisect each other
(a) In triangles AMB and CMD:
AB = CD (opposite sides of parallelogram)
∠ABM = ∠CDM (alternate angles, AB ∥ CD with diagonal BD as transversal)
∠AMB = ∠CMD (vertically opposite angles)
∴ △AMB ≡ △CMD (AAS).
(b) By CPCTC, AM = CM and BM = DM. ∴ The diagonals of a parallelogram bisect each other.
1.4 — Surveyor's check
In triangles PQR and P′Q′R′:
∠P = ∠P′ (given)
PR = P′R′ (given)
∠R = ∠R′ (given)
∴ △PQR ≡ △P′Q′R′ (AAS). (Two angles plus an included side — also called ASA in some textbooks; the lesson treats this as AAS.)
By CPCTC, PQ = P′Q′. The two river-spans are identical.
1.5 — Right-angled gables (RHS)
(a) Let the two gables be △ABC and △DEF, right-angled at B and E.
In triangles ABC and DEF:
∠B = ∠E = 90° (given right angles)
AC = DF = 5 m (hypotenuses — sloping rafters)
AB = DE = 3 m (vertical posts — corresponding legs)
∴ △ABC ≡ △DEF (RHS).
(b) By CPCTC, BC = EF, so the horizontal base of each gable is the same length (4 m, by the 3-4-5 triple).
2.1 — Explain your thinking (sample response)
(i) The friend's instinct is reasonable in that a well-drawn diagram does make a congruence "feel" obvious — visual recognition is a real mathematical skill. (ii) But the conclusion alone loses marks because in geometry the working is the answer: a marker awards marks for showing they can name matching parts with reasons, not just for the final equivalence symbol. A correct conclusion with no working typically earns 1 mark out of 3 or 4. (iii) A formal proof has four required parts in this order: (1) name the two triangles being compared, (2) list the matching parts with a reason in brackets after each one, (3) state the congruence with the test in brackets (e.g. SSS, SAS, AAS, RHS), and (4) use CPCTC for any unknown that follows. (iv) CPCTC is essential when the question asks you to use the congruence to deduce a specific side or angle — without it you have proved the triangles are identical but not actually answered the follow-up.
Marking: 1 for acknowledging visual instinct, 1 for the "working IS the answer" point, 1 for listing all four required parts in order including the phrase "test in brackets", 1 for explaining CPCTC's role.