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Problem 953

AMC 12 late, AIME early

Geometry Difficulty 4.8 Prove it Belarusian Mathematical Olympiad · Belarus

The point $X$ is marked inside the triangle $ABC$ . The circumcircles of the triangles $AXB$ and $AXC$ intersect the side $BC$ again at $D$ and $E$ respectively. The line $DX$ intersects the side $AC$ at $K$ , and the line $EX$ intersects the side $AB$ at $L$ .
Prove that $LK \parallel BC$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

First we prove that the points $A$ , $L$ , $X$ and $K$ lie on the same circle. Since the quadrilateral $ABDX$ is cyclic, $\angle AXK = \angle ABC$ . Similarly $\angle LXA = \angle BCA$ . Now from the equality $\angle KAL + \angle LXK = \angle CAB + \angle ABC + \angle BCA = 180^\circ$ it follows that the quadrilateral $ALXK$ is cyclic. Therefore $\angle KLX = \angle KAX$ .

Since the points $A$ , $X$ , $E$ and $C$ lie on the circle, $\angle DEX = \angle CAX$ . So $\angle KLX = \angle DEX$ and $LK \parallel BC$ .

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