Library / Slovenia — National Math Olympiad /54 of 122

Slovenia — National Math Olympiad

Geometry Difficulty 5.0 AIME, harder Prove it Slovenia

Let the bisectors of the angles $\angle CAD$ and $\angle ADB$ intersect the circumcircle of a cyclic quadrilateral $ABCD$ at $P$ and $Q$ . The lines $AP$ and $DQ$ intersect at $R$ , and the lines $CQ$ and $BP$ intersect at $S$ . Prove that the lines $PQ$ and $RS$ are perpendicular.

Solution

Since the quadrilateral $AQPD$ is cyclic, we have $\angle RPQ = \angle APQ = \angle ADQ$ . Since $DQ$ bisects the angle $ADB$ , we get $\angle ADQ = \angle QDB$ . Since the quadrilateral $QBPD$ is cyclic, we have $\angle QDB = \angle QPB = \angle QPS$ . Thus, $\angle RPQ = \angle QPS$ .

Similarly, since $QBPD$ is cyclic, we get $\angle PQR = \angle PQD = \angle PBD$ . The line $BP$ intersects the angle $\angle CBD$ , so $\angle PBD = \angle CBP$ . Since $QBCP$ is a cyclic quadrilateral, we have $\angle CBP = \angle CQP = \angle SQP$ . Thus $\angle PQR = \angle SQP$ .
The triangles $SPQ$ and $RPQ$ have congruent angles and a common side, so they are congruent. Hence, the quadrilateral $PRQS$ is a deltoid, which implies that $PQ \perp RS$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.