, 2017

Geometry Difficulty 8.6 Shortlist Prove it Hong Kong

In $\triangle ABC$ , $\angle B > \angle C$ . Let $D$ be the point on side $BC$ such that $\angle DAC = \frac{B-C}{2}$ . The circumcircle of $\triangle ACD$ meets side $AB$ again at $E$ . The circumcircle of $\triangle ABD$ meets side $AC$ again at $F$ . The internal angle bisector of $\angle BDE$ meets side $AB$ at $P$ . The internal angle bisector of $\angle CDF$ meets side $AC$ at $Q$ . Prove that $PQ$ and $AB$ are perpendicular.

Solutions — 2

Solution 1

Note that $\angle PDQ = 180^\circ - \angle BDP - \angle CDQ = 180^\circ - \frac{1}{2} \angle BDE - \frac{1}{2} \angle CDF = 180^\circ - \frac{A}{2} - \frac{A}{2} = 180^\circ - A$ . This implies $A, P, D, Q$ are concyclic. Hence, $\angle APQ = \angle ADQ = 180^\circ - \angle DAQ - \angle ACD - \angle CDQ = 180^\circ - \frac{B-C}{2} - C - \frac{A}{2} = 90^\circ$ . $\square$

Solution 2

From the concyclic points, we have $\angle DFC = \angle DBE$ and $\angle DCF = \angle DEB$ . These show $\triangle DFC$ and $\triangle DBE$ are directly similar. Thus, there exists a spiral similarity with centre $D$ mapping $\triangle DFC$ to $\triangle DBE$ . Note that the image of $Q$ is $P$ under this transformation since they are the corresponding points on sides $CF$ and $EB$ respectively. It follows that $\triangle DPQ$ and $\triangle DBF$ are directly similar. Then $\angle (PQ, AB) = \angle (PQ, BF) + \angle FBA = \angle (PD, BD) + (B - \angle CBF) = \frac{1}{2} \angle EDB + B - \angle DAC = \frac{A}{2} + B - \frac{B-C}{2} = 90^\circ$ . $\square$

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