Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.1 National Olympiad Prove it Canada

Problem:

Let five points on a circle be labelled $A$ , $B$ , $C$ , $D$ , and $E$ in clockwise order. Assume $A E = D E$ and let $P$ be the intersection of $A C$ and $B D$ . Let $Q$ be the point on the line through $A$ and $B$ such that $A$ is between $B$ and $Q$ and $A Q = D P$ . Similarly, let $R$ be the point on the line through $C$ and $D$ such that $D$ is between $C$ and $R$ and $D R = A P$ . Prove that $P E$ is perpendicular to $Q R$ .

Solution

Solution:

We are given $A Q = D P$ and $A P = D R$ . Additionally $\angle Q A P = 180^{\circ} - \angle B A C = 180^{\circ} - \angle B D C = \angle R D P$ , and so triangles $A Q P$ and $D P R$ are congruent. Therefore $P Q = P R$ . It follows that $P$ is on the perpendicular bisector of $Q R$ .

We are also given $A P = D R$ and $A E = D E$ . Additionally $\angle P A E = \angle C A E = 180^{\circ} - \angle C D E = \angle R D E$ , and so triangles $P A E$ and $R D E$ are congruent.

Therefore $P E = R E$ , and similarly $P E = Q E$ . It follows that $E$ is on the perpendicular bisector of $P Q$ .

Since both $P$ and $E$ are on the perpendicular bisector of $Q R$ , the result follows.

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