Midpoint theorem · 5 marks

(i) Complete the following proof of the Midpoint Theorem. Extend segment PQ beyond Q until point S. By how much should we extend PQ? It would be good to be able to prove that ∆APQ ≅ ∆CSQ. (Why?) Use this as a guide to specify the location of S and then complete this proof.
(ii) Give a similar proof of the converse of the Midpoint Theorem. Start by extending PQ up to a suitable point S.

(i) Complete the following proof of the Midpoint Theorem. Extend segment PQ beyond Q until point S. By how much should we extend PQ? It would be good to be able to prove that ∆APQ ≅ ∆CSQ. (Why?) Use this as a guide to specify the location of S and then complete this proof.
(ii) Give a similar proof of the converse of the Midpoint Theorem. Start by extending PQ up to a suitable point S.

Answer: (i) Extend PQ by its own length: QS = PQ. Then ∆APQ ≅ ∆CSQ (SAS), so CS = AP = PB and CS ‖ AB (∠PAQ = ∠SCQ, alternate). So PBCS is a parallelogram: PQ ‖ BC and PQ = ½PS = ½BC. The congruence is wanted because it moves AP across to CS, where it is equal and parallel to PB. (ii) For the converse (P the midpoint, PQ ‖ BC), extend PQ to S with PS = BC: PBCS is a parallelogram, so CS = PB = AP and CS ‖ AB; then ∆APQ ≅ ∆CSQ (ASA) gives AQ = QC (and PQ = QS = ½BC).

Step-by-step solution

Idea: Both proofs use the same picture: a copy of ∆APQ turned round Q onto ∆CSQ, which makes PBCS a parallelogram. In (i) we place S using what we know (Q is a midpoint); in (ii) we place S using PQ ‖ BC.

(i) Complete the following proof of the Midpoint Theorem. Extend segment PQ beyond Q until point S. By how much should we extend PQ? It would be good to be able to prove that ∆APQ ≅ ∆CSQ. (Why?) Use this as a guide to specify the location of S and then complete this proof.

Why the congruence helps: it would give CS = AP (= PB) and ∠SCQ = ∠PAQ, i.e. CS ‖ AB. Then PBCS would have a pair of opposite sides equal and parallel: a parallelogram, which is what gives PQ ‖ BC. For SAS we already have AQ = CQ and vertically opposite angles at Q, so we need QS = QP. So extend PQ by its own length: QS = PQ.1 mark
In ∆APQ and ∆CSQ: AQ = CQ (Q midpoint), ∠AQP = ∠CQS (vertically opposite), PQ = SQ (construction). So ∆APQ ≅ ∆CSQ (SAS).1 mark
Hence CS = AP = PB and ∠SCQ = ∠PAQ, which are alternate angles, so CS ‖ AB, i.e. CS ‖ PB. PBCS is a parallelogram (one pair of opposite sides equal and parallel). So PS ‖ BC and PS = BC, giving PQ ‖ BC and PQ = ½PS = ½BC.1 mark

Extend PQ by an equal length (QS = PQ); then ∆APQ ≅ ∆CSQ (SAS), PBCS is a parallelogram, and PQ ‖ BC, PQ = ½BC.

(ii) Give a similar proof of the converse of the Midpoint Theorem. Start by extending PQ up to a suitable point S.

Now P is the midpoint of AB and PQ ‖ BC (Q on AC). Extend PQ to S with PS = BC. PS ‖ BC and PS = BC, so PBCS is a parallelogram: CS = PB = AP and CS ‖ AB.1 mark
In ∆APQ and ∆CSQ: AP = CS, ∠PAQ = ∠SCQ and ∠APQ = ∠CSQ (alternate angles, AB ‖ CS). So ∆APQ ≅ ∆CSQ (ASA): AQ = CQ (Q is the midpoint of AC) and PQ = SQ = ½PS = ½BC.1 mark

Extend PQ to S with PS = BC; PBCS is a parallelogram, ∆APQ ≅ ∆CSQ (ASA), so AQ = QC and PQ = ½BC.

(i) Take QS = PQ; ∆APQ ≅ ∆CSQ makes PBCS a parallelogram, so PQ ‖ BC and PQ = ½BC. (ii) Take PS = BC; PBCS is a parallelogram, ∆APQ ≅ ∆CSQ, so Q is the midpoint of AC.

Answer to write in the exam

(i)

Produce PQ to S with QS = PQ

∆APQ ≅ ∆CSQ (SAS: AQ = CQ, ∠AQP = ∠CQS, PQ = SQ)

⇒ CS = AP = PB and ∠SCQ = ∠PAQ ⇒ CS ‖ PB (alternate angles)

⇒ PBCS is a parallelogram ⇒ PS ‖ BC, PS = BC

∴ PQ ‖ BC and PQ = ½BC

(ii)

Produce PQ to S with PS = BC; PS ‖ BC ⇒ PBCS is a parallelogram ⇒ CS = PB = AP, CS ‖ AB

∆APQ ≅ ∆CSQ (ASA: ∠PAQ = ∠SCQ, AP = CS, ∠APQ = ∠CSQ)

∴ AQ = QC and PQ = QS = ½BC

Common mistakes that cost marks

Extending PQ by an arbitrary amount: only QS = PQ gives the SAS congruence in (i).
In (ii), choosing QS = PQ: that assumes Q is already known to be special. Use PS = BC instead, which only uses PQ ‖ BC.
Calling PBCS a parallelogram from CS = PB alone; the parallel condition is also needed.

How this can come in the exam

Short answer (2 marks)

In the proof of the Midpoint Theorem, PQ is produced to S with QS = PQ. If PQ = 3.4 cm, find PS and BC.

Show answer

PS = 2PQ = 6.8 cm (1 mark). PBCS is a parallelogram, so BC = PS = 6.8 cm (1 mark).

Try one yourself

In the proof above, why is ∠SCQ = ∠PAQ enough to show CS ‖ AB?