Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.8 National olympiad Prove it

[^6]16. G4 (AUS-ARM) ${ }^{\mathrm{IMO} 2} N$ is an arbitrary point on the bisector of $\angle B A C$ . $P$ and $O$ are points on the lines $A B$ and $A N$ , respectively, such that $\measuredangle A N P=90^{\circ}=\measuredangle A P O$ . $Q$ is an arbitrary point on $N P$ , and an arbitrary line through $Q$ meets the lines $A B$ and $A C$ at $E$ and $F$ respectively. Prove that $\measuredangle O Q E=90^{\circ}$ if and only if $Q E=Q F$ .

Solution

16. First, assume that $\angle O Q E=90^{\circ}$ . Extend $P N$ to meet $A C$ at $R$ . Then $O E P Q$ and $O R F Q$ are cyclic quadrilaterals; hence we have $\angle O E Q=$ $\angle O P Q=\angle O R Q=\angle O F Q$ . It follows that $\triangle O E Q \cong \triangle O F Q$ and $Q E=Q F$ . Now suppose $Q E=Q F$ . Let $S$ be the point symmetric to $A$ with respect to $Q$ , so that the quadrilateral $A E S F$ is a parallelogram. Draw the line $E^{\prime} F^{\prime}$ through $Q$ so that $\angle O Q E^{\prime}=90^{\circ}$ and $E^{\prime} \in A B$ , $F^{\prime} \in A C$ . By the first part $Q E^{\prime}=$ ! $Q F^{\prime}$ ; hence $A E^{\prime} S F^{\prime}$ is also a parallelogram. It follows that $E \equiv E^{\prime}, F \equiv F^{\prime}$ , and $\angle O Q E=90^{\circ}$ .

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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.