Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it Bulgaria

Problem:

The points $P$ and $Q$ lie in the interior of $\triangle ABC$ , $\angle ACP = \angle BCQ$ and $\angle CAP = \angle BAQ$ . The feet of the perpendiculars from $P$ to the lines $BC$ , $CA$ and $AB$ are denoted by $D$ , $E$ and $F$ , respectively. Prove that if $\angle DEF = 90^{\circ}$ , then $Q$ is the orthocenter of $\triangle BDF$ .

Solution

Solution:

We have $\angle BCQ = \angle ACP = \angle EDP$ . Since $PD \perp BC$ , it follows that $ED \perp CQ$ . Analogously we have $AQ \perp EF$ . Since $\angle DEF = 90^{\circ}$ , we conclude that $\angle AQC = 90^{\circ}$ as well. Then $\triangle QCD \sim \triangle ACP$ , because $\angle QCD = \angle ACP$ and we get
$\frac{DC}{QC} = \frac{PC \cos \angle PCD}{AC \cos \angle ACQ} = \frac{PC}{AC}$
Therefore $\angle DQC = \angle PAC = \angle PFE$ . Since $CQ \parallel EF\ (\perp ED)$ , it follows that $DQ \parallel PF$ , i.e.

$DQ \perp AB$ .
Using the same arguments we prove that $FQ \perp BC$ and hence $Q$ is the orthocenter of $\triangle BDF$ .

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