Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.0 National olympiad Prove it

II. (25 points) In an acute triangle $\triangle ABC$ , $M$ is the foot of the perpendicular from $A$ to $BC$ , with $P$ and $Q$ being the feet of the perpendiculars from $B$ and $C$ to $AC$ and $AB$ , respectively. If $AD \perp PQ$ , and $AD$ intersects the circumcircle of $\triangle ABC$ at $D$ , prove that $PQ=\frac{2 S_{\triangle ABC}}{AD}$ .

Solution

Connect $C D$ .
$\because M P \perp A B, M Q \perp A C,$
$\therefore A, P, M, Q$ are concyclic, and $A M$ is the diameter of this circle. By the Inscribed Angle Theorem,
$\begin{array}{l} P Q=A M \cdot \sin 4. \\ \because A D \perp P, A C \perp M Q, \\ \therefore \angle 2=\angle 3 . \end{array}$

Since $\angle 1=\angle 2$ ,
we have $\angle 1=\angle 3$ .
Also, $\angle B=\angle D$ ,
$\therefore \triangle A B M \sim \triangle A D C$ .
$\therefore A B \cdot A C=A D \cdot A M \text {, }$

which means $\frac{1}{2} A B \cdot A C \cdot \sin A=\frac{1}{2} A D \cdot A M \cdot \sin A$ .
From (1) and (2), we get $S_{\triangle A B C}=\frac{1}{2} A D \cdot P Q$ , i.e., $P Q=\frac{2 S_{\triangle A B C}}{A D}$ .

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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.