Competition problem — Geometry — Olympiad Maths Prep

Geometry Difficulty 6.5 National olympiad Prove it China

As shown below, let $ABCD$ be a cyclic quadrilateral such that the diagonals $AC$ and $BD$ are perpendicular with intersection point $E$ . Point $F$ is on the side $AD$ , the ray $FE$ meets the circumscribed circle of $ABCD$ at point $P$ . Point $Q$ is on the segment $PE$ such that $PQ \cdot PF = PE^2$ . The line through $Q$ perpendicular to $AD$ meets $BC$ at $R$ . Show that $RP = RQ$ .

Solution

Proof: Let $EX \parallel AF$ , intersecting $AP$ at $X$ , and $DP$ at $Y$ . Extend $XQ$ and $YQ$ to intersect $BC$ at $S$ and $T$ , respectively.
From $\frac{PQ}{PE} = \frac{PE}{PF} = \frac{PX}{PA}$ , we get $XQ \parallel AE$ , similarly $YQ \parallel DE$ .
Since $\angle EXS = \angle AEX = \angle DAC = \angle EBS$ , we have $X, E, S, B$ are concyclic.
From $\angle PXE = \angle PAD = \angle PBE$ , we find $X, E, P, B$ are concyclic, hence $X, E, S, P, B$ are concyclic. Similarly, $Y, E, T, P, C$ are concyclic.
Since $\angle PQT = \angle PEB = \angle PST$ , we have $P, S, Q, T$ are concyclic. Given $SQ \parallel CE, TQ \parallel BE$ , and $CE \perp BE$ , we get $SQ \perp TQ$ .
Since $\angle RQS = 90^\circ - \angle QXY = 90^\circ - \angle QTS = \angle RSQ$ , it is known that $R$ is the center of circle $\odot PSQT$ , thus $RP = RQ$ . $\square$

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