Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.2 National olympiad Prove it

In quadrilateral $A B C D$ , $E$ is the intersection of the diagonals. A line through $E$ , not equal to $A C$ or $B D$ , intersects $A B$ at $P$ and $B C$ at $Q$ . The circle that is tangent to $P Q$ at $E$ and also passes through $D$ , intersects the circumcircle of $A B C D$ again at point $R$ . Prove that $B, P, R$ and $Q$ lie on a circle.
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We consider the configuration where $P$ lies internally on $A B$ and $Q$ lies externally on $B C$ ; other configurations proceed analogously.

Solution

For the cyclic quadrilateral $D B C R$ , we have $\angle Q C R=180^{\circ}-\angle B C R=\angle B D R$ . Since $E Q$ is tangent to the circle through $E, D$ , and $R$ , we have $\angle B D R=\angle E D R=\angle Q E R$ , so $\angle Q C R=\angle Q E R$ . This means that $E R Q C$ is a cyclic quadrilateral. Similarly, $E P A R$ is also a cyclic quadrilateral.

We now find $\angle P R Q=\angle P R E+\angle E R Q=\angle P A E+180^{\circ}-\angle E C Q=\angle B A E+\angle E C B$ . Due to the angle sum in $\triangle A B C$ , this is equal to $180^{\circ}-\angle A B C=180^{\circ}-\angle P B Q$ . Therefore, $\angle P R Q=180^{\circ}-\angle P B Q$ , and it follows that $B P R Q$ is a cyclic quadrilateral.

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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.