In a convex quadrilateral $ABCD$, $\angle ABC = \angle CDA$. Let $X \neq C$ be the intersection of the circumcircle of $\triangle BCD$ and circle with diameter $AC$. Prove that the tangent to the circumcircle of $\triangle BCD$ at $X$, the tangent to the circumcircle of $\triangle ABD$ at $A$ concur on $BD$.
Problem
Source: 2025 Korea Winter Program Practice Test P5
Tags: geometry, circumcircle
Acorn-SJ
19.01.2025 19:09
A very minor detail, but the original problem mentions that the quadrilateral $ABCD$ is convex. Also I don't think this problem is anywhere near trivial.
khina
19.01.2025 21:13
Should that be the tangent to the circumcircle of $ABD$ at $A$? Otherwise the problem implies that the circumcircles of $ACD$ and $ABC$ are tangent at $A$, which is absurd...?
khina
19.01.2025 21:18
Anyways, the problem is correct if $ACD$ is replaced with $ABD$ at the end.
It suffices to prove that $ABXD$ is harmonic. Inverting at $A$, it suffices to prove that after inversion $XB = XD$. But after inversion we find that $A$ lies on an angle bisector of $\angle{BCD}$, $XC \perp CA$ (so $X$ also lies on an angle bisector), and $X$ also lies on $(BCD)$. This implies $XB = XD$ as desired.
Acorn-SJ
21.01.2025 11:27
@above Can we say that a quarilateral is harmonic when it is not cyclic? I have only seen cases where harmonic quadrilaterals were cyclic.
It suffices to prove $\frac{BX^2}{XD^2}=\frac{BA^2}{AD^2}$. This translates to showing that $BXDJ$ is harmonic where $J$ is the intersection of $AC$ and $(BCD).$ This becomes apparent under $A$-pencil.