Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.4 AIME, harder Prove it

II. (25 points) In $\triangle ABC$ , from point $A$ draw perpendiculars to the angle bisectors of $\angle B$ and $\angle C$ , with the feet of the perpendiculars being $P$ and $Q$ respectively; from point $B$ draw a perpendicular to the angle bisector of $\angle C$ , with the foot of the perpendicular being $E$ ; from point $C$ draw a perpendicular to the angle bisector of $\angle B$ , with the foot of the perpendicular being $F$ . Prove that points $P$ , $Q$ , $E$ , and $F$ are concyclic.

Solution

Extend $A P$ and $A Q$ to intersect $B C$ at $M$ and $N$ respectively.
$\because B P \text{ bisects } \angle A B C, A P \perp B P,$
$\therefore A P=P M \text{. }$

Similarly, $A Q=Q N$ .
$\therefore P Q \parallel B C, \angle Q P B=\angle P B C \text{. }$

Connect $E F$ .
$\because \angle B E C=90^{\circ}=\angle C F B \text{, }$
$\therefore B, C, F, E$ are concyclic.
Thus $\angle C E F=\angle F B C$ .
Therefore, $\angle Q P B=\angle C E F$ .
$\therefore P, Q, E, F$ are concyclic.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.