Prove the internal angle bisector theorem, namely that the internal bisector of an angle of a triangle divides the side opposite the angle into segments proportional to the sides adjacent to the angle.

The bisector of the exterior angle $\widehat A$ of the triangle ABC meets (BC) at P. The bisector of the interior angle $\widehat B$ meets [AC] at Q. Given that (PQ) meets [AB] at R, use Menelaus’ theorem to prove that (CR) bisects the angle ${\rm{A}}\widehat {\rm{C}}{\rm{B}}$ .

EITHER

let [AD] bisect A, draw a line through C parallel to (AD) meeting (AB) at E     M1

then ${\rm{B}}\widehat {\rm{A}}{\rm{D}} = {\rm{A}}\widehat {\rm{E}}{\rm{C}}$ and ${\rm{D}}\widehat {\rm{A}}{\rm{C}} = {\rm{A}}\widehat {\rm{C}}{\rm{E}}$     A1

since ${\rm{B}}\widehat {\rm{A}}{\rm{D}} = {\rm{D}}\widehat {\rm{A}}{\rm{C}}$ it follows that ${\rm{A}}\widehat {\rm{E}}{\rm{C}} = {\rm{A}}\widehat {\rm{C}}{\rm{E}}$     A1

triangle AEC is therefore isosceles and ${\rm{AE}} = {\rm{AC}}$     A1

since triangles BAD and BEC are similar

$\frac{{{\rm{BD}}}}{{{\rm{DC}}}} = \frac{{{\rm{AB}}}}{{{\rm{AE}}}} = \frac{{{\rm{AB}}}}{{{\rm{AC}}}}$     M1A1

OR

$\frac{{{\rm{AB}}}}{{\sin \beta }} = \frac{{{\rm{BD}}}}{{\sin \alpha }}$     M1A1

$\frac{{{\rm{AC}}}}{{\sin (180 - \beta )}} = \frac{{{\rm{DC}}}}{{\sin \alpha }}$     M1A1

$\sin \beta  = \sin (180 - \beta )$     R1

$\Rightarrow \frac{{{\rm{AB}}}}{{{\rm{BD}}}} = \frac{{{\rm{AC}}}}{{{\rm{DC}}}}$

$\Rightarrow \frac{{{\rm{BD}}}}{{{\rm{DC}}}} = \frac{{{\rm{AB}}}}{{{\rm{AC}}}}$     A1

[6 marks]

using the angle bisector theorem,     M1

$\frac{{{\rm{AQ}}}}{{{\rm{QC}}}} = \frac{{{\rm{AB}}}}{{{\rm{BC}}}}$ and $\frac{{{\rm{BP}}}}{{{\rm{PC}}}} = \frac{{{\rm{AB}}}}{{{\rm{BC}}}}$     A1

using Menelaus’ theorem with (PR) as transversal to triangle ABC     M1

$\frac{{{\rm{BR}}}}{{{\rm{AR}}}} \times \frac{{{\rm{AQ}}}}{{{\rm{QC}}}} \times \frac{{{\rm{PC}}}}{{{\rm{BP}}}} = ( - )1$     A1

substituting the above results,     M1

$\frac{{{\rm{BR}}}}{{{\rm{AR}}}} \times \frac{{{\rm{AB}}}}{{{\rm{BC}}}} \times \frac{{{\rm{AC}}}}{{{\rm{AB}}}} = ( - )1$     A1

giving

$\frac{{{\rm{BR}}}}{{{\rm{AR}}}} = \frac{{{\rm{BC}}}}{{{\rm{AC}}}}$     A1

[CR] therefore bisects angle C by (the converse to) the angle bisector theorem     R1AG

[8 marks]