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Saudi Arabian Mathematical Competitions

Geometry Difficulty 6.5 National Olympiad Prove it Saudi Arabia

Let $A$ be a point outside the circle $\omega$ . Two points $B, C$ lie on $\omega$ such that $A B, A C$ are tangent to $\omega$ . Let $D$ be any point on $\omega$ ( $D$ is neither $B$ nor $C$ ) and $M$ the foot of perpendicular from $B$ to $C D$ . The line through $D$ and the midpoint of $B M$ meets $\omega$ again at $P$ . Prove that $A P \perp C P$ .

Solution

Let $O$ be the center of $\omega$ and $Q$ the intersection point of $C O$ and $\omega$ .
Since $Q D$ and $B M$ are perpendicular to $C D$ then $B M \parallel Q D$ . Because $N$ is the midpoint of $B M$ , which implies that $D(M, B, N, Q) = -1$ .
Consider this harmonic quartet with circle $\omega$ , one has $C P B Q$ is a harmonic quadrilateral. This implies that $A, P, Q$ are collinear and $A P \perp C P$ .

$\square$

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