The points P, Q and R, lie on the sides [AB], [AC] and [BC], respectively, of the triangle ABC. The lines (AR), (BQ) and (CP) are concurrent.

Use Ceva’s theorem to prove that [PQ] is parallel to [BC] if and only if R is the midpoint of [BC].

suppose R is the midpoint of BC     M1

Note:     The first mark is for initiating a relevant discussion for “if” or “only if” by Ceva’s theorem.

\(\frac{{{\text{AP}}}}{{{\text{PB}}}} \times \frac{{{\text{BR}}}}{{{\text{RC}}}} \times \frac{{{\text{CQ}}}}{{{\text{QA}}}} = 1\)    A1

\( \Rightarrow \frac{{{\text{AP}}}}{{{\text{PB}}}} = \frac{{{\text{AQ}}}}{{{\text{QC}}}}\) or equivalent     A1

\( \Rightarrow \frac{{{\text{PB}}}}{{{\text{AP}}}} + 1 = \frac{{{\text{QC}}}}{{{\text{AQ}}}} + 1\)    (M1)

\( \Rightarrow \frac{{{\text{AP}} + {\text{PB}}}}{{{\text{AP}}}} = \frac{{{\text{AQ}} + {\text{QC}}}}{{{\text{AQ}}}}\)

\( \Rightarrow \frac{{{\text{AB}}}}{{{\text{AP}}}} = \frac{{{\text{AC}}}}{{{\text{AQ}}}}\)    A1

\( \Rightarrow \) triangles APQ and ABC are similar with common base angles     R1

so PQ is parallel to BC     AG

statement of the converse     A1

the argument is reversible     R1AG

[8 marks]

This was again a question which a significant number of students were unable to start. For those who did start only a small number understood the significance of “if and only if” meaning that wholly correct answers were not often seen.