Auswahlklausur

Geometry Difficulty 8.2 Shortlist Prove it Germany

Let $ABCD$ be a convex quadrilateral. The diagonal $BD$ bisects the angle $\measuredangle CBA$ . The circumcircle of triangle $ABC$ intersects the segments $\overline{CD}$ and $\overline{DA}$ at the interior points $P$ and $Q$ , respectively. The line through the point $D$ parallel to the line $AC$ intersects the lines $BA$ and $BC$ at the points $R$ and $S$ , respectively. Prove that the four points $P, Q, R$ and $S$ lie on a common circle.

Solution

Step 1: Since $\measuredangle BRD = \measuredangle BAC = \measuredangle BPC = \pi - \measuredangle DPB$ , $BPDR$ is a cyclic quadrilateral. Analogously one shows that $BSDQ$ is also a cyclic quadrilateral.

Step 2: Let $X$ be defined as the intersection point of $BD$ with the circumcircle of triangle $ABC$ . Then $\measuredangle DPX = \pi - \measuredangle XPC = \measuredangle CBX = 1/2 \measuredangle CBA$ holds. Since by Step 1 we also have $\measuredangle DPR = \measuredangle DBR = 1/2 \measuredangle DBA$ , $X$ lies on $PR$ . Analogously one shows that $X$ also lies on $QS$ .

Step 3: We combine the findings and compute (using the power of a point theorem) $|PX| \cdot |XR| = |BX| \cdot |XD| = |QX| \cdot |XS|$ . Now the claim follows by the converse of the power of a point theorem.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from de; metadata (topic, difficulty) added by this project.