D, P, Q are collinear

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D, P, Q are collinear

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Source: Korea 2002 : P5

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#1 h Jun 1, 2014, 2:56 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an acute triangle and let $\omega$ be its circumcircle. Let the perpendicular line from $A$ to $BC$ meet $\omega$ at $D$ . Let $P$ be a point on $\omega$ , and let $Q$ be the foot of the perpendicular line from $P$ to the line $AB$ . Prove that if $Q$ is on the outside of $\omega$ and $2\angle QPB = \angle PBC$ , then $D,P,Q$ are collinear.

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#2 h Jun 1, 2014, 7:17 AM • 2 Y

Y by Adventure10, Mango247

Note that our claim is true only if $AB<AC$ . If $AB\ge AC$ , then $\angle D=90+\angle B-\angle C\le 90$ , which means that $\angle D$ is acute and $Q$ lies on segment $AB$ or inside $w$ .

Let $PQ\cap BC=E$ , hence $\angle QPB=\angle PBC-\angle QPB=\angle BEQ=90-\angle B=180-\angle BPD\Rightarrow Q,P,D$ are collinear.

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#3 h Oct 22, 2014, 1:47 AM • 1 Y

Y by Adventure10

Simple angle chasing yields $\angle BQP=\angle ABC$ , so letting $X=AD\cap BC$ we get $\triangle BQP\sim\triangle BXA\implies \angle BQX\sim\triangle BPA$ . Then since $BQDX$ is cyclic we have $\angle BDP=\pi-\angle BAP=\pi-\angle BXQ=\pi-\angle BDQ,$ establishing the desired collinearity.

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