AoPS - Problem

Source: Moroccan TST 2019 P6

Tags: geometry, circumcircle

DrAymeinstein

23.08.2024 19:03

Let $ABC$ be a triangle. The tangent in $A$ of the circumcircle of $ABC$ cuts the line $(BC)$ in $X$. Let $A'$ be the symetric of $A$ by $X$ and $C'$ the symetric of $C$ by the line $(AX)$ Prove that the points $A, C', A'$ and $B$ are concyclic.

let $D$ be the reflection of $C$ in $X$. claim 1: $A'AC'D$ is cyclic. Proof: by reflections, $AX\parallel C'D$ amd $ACA'C'$ is a parallelogram. Thus we have $\measuredangle AA'D=\measuredangle A'AC=\measuredangle C'AA'$ so $A'AC'D$ is an isosceles trapezoid. claim 2: $ABC'D$ is cyclic. Proof: note that $\measuredangle AC'D=\measuredangle AC'X+\measuredangle XC'D=\measuredangle XCA+\measuredangle C'XA=\measuredangle BAX+\measuredangle AXB=\measuredangle ABC=\measuredangle ABD$ which implies $ABC'D$ cyclic. these 2 claims imply the desired conclusion.