ABCD is a quadrilateral. (AD) and (BC) intersect at F and (AB) and (CD) intersect at H. (DB) and (CA) intersect (FH) at G and E respectively. This is shown in the diagram below.

Prove that \(\frac{{{\rm{HG}}}}{{{\rm{GF}}}} = - \frac{{{\rm{HE}}}}{{{\rm{EF}}}}\) .

in \(\Delta {\rm{HFD}}\) , [HA], [FC] and [DG] are concurrent at B     M1

so, \(\frac{{{\rm{HG}}}}{{{\rm{GF}}}} \times \frac{{{\rm{FA}}}}{{{\rm{AD}}}} \times \frac{{{\rm{DC}}}}{{{\rm{CH}}}} = 1\) by Ceva’s theorem     A1R1

in \(\Delta {\rm{HFD}}\) , with CAE as transversal,     M1

\(\frac{{{\rm{HE}}}}{{{\rm{EF}}}} \times \frac{{{\rm{FA}}}}{{{\rm{AD}}}} \times \frac{{{\rm{DC}}}}{{{\rm{CH}}}} = - 1\) by Menelaus’ theorem     A1R1

therefore, \(\frac{{{\rm{HG}}}}{{{\rm{GF}}}} = - \frac{{{\rm{HE}}}}{{{\rm{EF}}}}\)     AG

[6 marks]

This was not a difficult question but again too many candidates often left gaps in their solutions perhaps thinking that what they were doing was obvious and needed no written support