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Hong Kong Team Selection Test 1

Geometry Difficulty 8.3 Shortlist Prove it Hong Kong

Given $\triangle ABC$ with $AB < AC$ , let $AD$ be the bisector of $\angle BAC$ with $D$ on the side $BC$ . Let $\Gamma$ be a circle passing through $A$ and $D$ which is tangent to $BC$ at $D$ . Suppose $\Gamma$ cuts the side $AB$ again at $E \neq A$ . The tangent to the circumcircle of $\triangle BDE$ at $D$ intersects $\Gamma$ again at $F \neq D$ . Let $P$ be the intersection point of the segments $EF$ and $AC$ . Prove that $PD$ is perpendicular to $BC$ .

Solution

Let $Q \neq A$ be the second intersection point of $\Gamma$ and $AC$ . As
$\angle EDB = \angle EAD = \angle DAQ = \angle DEQ,$
we have $EQ // BC$ . Also, we have $AF // BC$ since
$\angle EAF = 180^\circ - \angle FDE = 180^\circ - \angle DBE.$

This shows $EQ // AF$ , and hence $AEQF$ is an isosceles trapezoid. Thus, the intersection point $P$ of the diagonals lies on the perpendicular bisector of $EQ$ . Note that $D$ also lies on the perpendicular bisector of $EQ$ because $AD$ bisects $\angle EAQ$ and $A$ , $E$ , $D$ , $Q$ are concyclic. Therefore, $PD \perp EQ$ , which implies $PD \perp BC$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.