The diagram shows the line \(l\) meeting the sides of the triangle ABC at the points D, E and F. The perpendiculars to \(l\) from A, B and C meet \(l\) at G, H and I.

  (i)     State why \(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} = \frac{{{\rm{AG}}}}{{{\rm{HB}}}}\) .

  (ii)     Hence prove Menelaus’ theorem for the triangle ABC.

  (iii)     State and prove the converse of Menelaus’ theorem.

A straight line meets the sides (PQ), (QR), (RS), (SP) of a quadrilateral PQRS at the points U, V, W, X respectively. Use Menelaus’ theorem to show that\[\frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} = 1.\]

(i)     Because the triangles AGF and BHF are similar.     R1

(ii)     It follows (by cyclic rotation or considering similar triangles) that

\(\frac{{{\rm{BD}}}}{{{\rm{DC}}}} = \frac{{{\rm{BH}}}}{{{\rm{IC}}}}\)     A1

and \(\frac{{{\rm{CE}}}}{{{\rm{EA}}}} = \frac{{{\rm{CI}}}}{{{\rm{GA}}}}\)     A1

Multiplying these three results gives Menelaus’ Theorem, i.e.

\(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} \times \frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} = \frac{{{\rm{AG}}}}{{{\rm{HB}}}} \times \frac{{{\rm{BH}}}}{{{\rm{IC}}}} \times \frac{{{\rm{CI}}}}{{{\rm{GA}}}}\)     M1A1

\( = \frac{{{\rm{AG}}}}{{{\rm{GA}}}} \times \frac{{{\rm{BH}}}}{{{\rm{HB}}}} \times \frac{{{\rm{CI}}}}{{{\rm{IC}}}} =  - 1\)     M1A1

(iii)     The converse states that if D, E, F are points on the sides (BC), (CA), (AB) of a triangle such that

\(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} \times \frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} = - 1\)

then D, E, F are collinear.     A1

To prove this result, let D, E, F′ be collinear points on the three sides so that, using the above theorem,     M1

\(\frac{{{\rm{AF'}}}}{{{\rm{F'B}}}} \times \frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} = - 1\)    A1

Since \(\frac{{{\rm{AF}}}}{{{\rm{FB}}}} \times \frac{{{\rm{BD}}}}{{{\rm{DC}}}} \times \frac{{{\rm{CE}}}}{{{\rm{EA}}}} = - 1\)     M1

\(\frac{{{\rm{AF'}}}}{{{\rm{F'B}}}} = \frac{{{\rm{AF}}}}{{{\rm{FB}}}}\)     A1

and \({\rm{F = F'}}\) which proves the converse.     R1

[13 marks]

Draw the diagonal PR and let it cut the line at the point Y.     M1

Apply Menelaus’ Theorem to the triangle PQR. Then,

\(\frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RY}}}}{{{\rm{YP}}}} =  - 1\)     M1A1

Now apply the theorem to triangle PRS.

\(\frac{{{\rm{PY}}}}{{{\rm{YR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} =  - 1\)     A1

\(\frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RY}}}}{{{\rm{YP}}}} \times \frac{{{\rm{PY}}}}{{{\rm{YR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} =  - 1 \times  - 1\)     M1

\( \Rightarrow \frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} \times \frac{{{\rm{PY}}}}{{{\rm{YP}}}} \times \frac{{{\rm{RY}}}}{{{\rm{YR}}}} = 1\)    A1

\( \Rightarrow \frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} \times ( - 1) \times ( - 1) = 1\)     (M1)

\( \Rightarrow \frac{{{\rm{PU}}}}{{{\rm{UQ}}}} \times \frac{{{\rm{QV}}}}{{{\rm{VR}}}} \times \frac{{{\rm{RW}}}}{{{\rm{WS}}}} \times \frac{{{\rm{SX}}}}{{{\rm{XP}}}} = 1\)     AG

[7 marks]