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Geometry Difficulty 8.7 Shortlist Prove it Hong Kong

Let $ABCD$ be a cyclic quadrilateral and $E$ be the intersection of $AC$ and $BD$ . $P$ and $Q$ are two points on $AC$ such that the points $A$ , $E$ , $Q$ , $P$ , $C$ lie on the same straight line in this order, and that $BP$ bisects $\angle ABC$ whereas $DQ$ bisects $\angle ADC$ . If $AE = 4$ , $EQ = 2$ and $QP = 3$ , find the length of $PC$ .

Solution

Note that $\sin \angle DAB = \sin \angle DCB$ as the two angles are supplementary. Using $[XYZ]$ to denote the area of $XYZ$ , we have
$\frac{AE}{EC} = \frac{[DAB]}{[DCB]} = \frac{\frac{1}{2} \cdot AD \cdot AB}{\frac{1}{2} \cdot DC \cdot BC} = \frac{AB}{BC} \cdot \frac{AD}{DC} \quad (1)$
By the angle bisector theorem, we have
$\frac{AB}{BC} = \frac{AP}{PC} \quad \text{and} \quad \frac{AD}{DC} = \frac{AQ}{QC}$
It follows from (1) that
$\frac{AE}{EC} = \frac{AP}{PC} \cdot \frac{AQ}{QC}$
Setting $PC = x$ and using the given side lengths, this becomes
$\frac{4}{5+x} = \frac{9}{x} \cdot \frac{6}{3+x}$
Solving gives $x = 15$ or $x = -\frac{9}{2}$ , and of course the latter is rejected.

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