Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.0 National olympiad Prove it

Let $A B C$ be a triangle with acute angles and let $D$ be a point inside the triangle $A B C$ . The lines $(A D)$ and $(B D)$ intersect the circumcircle of triangle $A B C$ again at points $A_{1}$ and $B_{1}$ , respectively. The circumcircle of triangle $B_{1} D A$ intersects the line $(A C)$ at point $P$ . The circumcircle of triangle $A_{1} B D$ intersects the line $(B C)$ at point $Q$ .
Prove that the quadrilateral $C P D Q$ is a parallelogram.

Solution

A hunt for angles of straight lines indicates that

$(Q D, C B)=(Q D, Q B)=\left(A_{1} D, A_{1} B\right)=\left(A_{1} A, A_{1} B\right)=(C A, C B),$

which means that the lines $(Q D)$ and $(A C)$ are parallel to each other. Similarly, we can prove that

$(P D, C A)=(P D, P A)=\left(B_{1} D, B_{1} A\right)=\left(B_{1} B, B_{1} A\right)=(C B, C A)$

which means that the lines $(P D)$ and $(B C)$ are parallel to each other. Therefore, the quadrilateral $C P D Q$ is a parallelogram.
!

Examiner's Comment The exercise is very well solved.

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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.