(a)     Show that, with the usual convention for the signs of lengths in a triangle, \(\frac{{{\text{BD}}}}{{{\text{DC}}}} =  - \frac{{{\text{BG}}}}{{{\text{GC}}}}\).

(b)     The lines \({\text{(FD), (CA)}}\) meet at the point \({\text{H}}\) and the lines \({\text{(DE), (AB)}}\) meet at the point \({\text{I}}\). Show that the points \({\text{G, H, I}}\) are collinear.

(a)     applying Ceva’s theorem to triangle \({\text{ABC}}\),

\(\frac{{{\text{CD}}}}{{{\text{DB}}}} \times \frac{{{\text{AE}}}}{{{\text{EC}}}} \times \frac{{{\text{BF}}}}{{{\text{FA}}}} = 1\)     M1A1

applying Menelaus’ theorem to triangle \({\text{ABC}}\) with transversal \({\text{(GFE)}}\),

\(\frac{{{\text{BG}}}}{{{\text{GC}}}} \times \frac{{{\text{CE}}}}{{{\text{EA}}}} \times \frac{{{\text{AF}}}}{{{\text{FB}}}} -  = 1\)     M1A1

multiplying the two equations,     M1

\(\frac{{{\text{CD}}}}{{{\text{DB}}}} \times \frac{{{\text{BG}}}}{{{\text{GC}}}} =  - 1\)     A1

so that \(\frac{{{\text{BD}}}}{{{\text{DC}}}} =  - \frac{{{\text{BG}}}}{{{\text{GC}}}}\)     AG

[6 marks]

(b)     similarly

\(\frac{{{\text{CE}}}}{{{\text{EA}}}} =  - \frac{{{\text{CH}}}}{{{\text{HA}}}}\)     M1A1

and \(\frac{{{\text{AF}}}}{{{\text{FB}}}} =  - \frac{{{\text{AI}}}}{{{\text{IB}}}}\)     A1

multiplying the three results,

\(\frac{{{\text{BD}}}}{{{\text{DC}}}} \times \frac{{{\text{CE}}}}{{{\text{EA}}}} \times \frac{{{\text{AF}}}}{{{\text{FB}}}} =  - \frac{{{\text{BG}}}}{{{\text{GC}}}} \times \frac{{{\text{CH}}}}{{{\text{HA}}}} \times \frac{{{\text{AI}}}}{{{\text{IB}}}}\)     M1A1

by Ceva’s theorem, as shown previously, the left hand side is equal to \(1\), therefore so is the right hand side     R1

that is \(\frac{{{\text{BG}}}}{{{\text{GC}}}} \times \frac{{{\text{CH}}}}{{{\text{HA}}}} \times \frac{{{\text{AI}}}}{{{\text{IB}}}} =  - 1\)     A1

it follows from the converse to Menelaus’ theorem that \({\text{G, H, I}}\) are collinear     R1

[8 marks]