Let $ABCD$ be a quadrilateral inscribed in a circle $\Gamma$ such that $AB=BC=CD$. Let $M$ and $N$ be the midpoints of $AD$ and $AB$ respectively. The line $CM$ meets $\Gamma$ again at $E$. Prove that the tangent at $E$ to $\Gamma$, the line $AD$ and the line $CN$ are concurrent.
Problem
Source: 2023 Hong Kong TST 3 (CHKMO)
Tags: geometry
LLL2019
04.12.2022 05:29
I found this familiar during the test. Nevertheless, I couldn't find the synthetic solution, and I length chased it (compute $XE^2$ by LoC twice on $XEM$ and $DMC,$ where $X=AD\cap CN.$). In fact, this is equivalent to JMO 2011/5.
bxiao31415
21.11.2023 17:15
First, note that $(D,A;E,B)=C(D,A;M,\infty)=-1$, so $ABDE$ is harmonic. This implies that the tangents at $B$ and $E$ to $\Gamma$ and $AD$ are concurrent. Hence it suffices to show that $AD$, the tangent at $B$, and $CN$ are concurrent. Let $P$ be the intersection of $AD$ and $CN$. $BN=NA$ and the fact that $CB$ and $AP$ are parallel tell us $\bigtriangleup CBN \equiv \bigtriangleup PAN$, so $CB = PA$. This then gives $\bigtriangleup ABP \equiv \bigtriangleup BAC$ as $\angle{CBA}=\angle{PAB}$. Hence $\angle{PBA}=\angle{CAB}=\angle{BCA}$ as $BA=BC$. Therefore, $PB$ is tangent to $\Gamma$, so $P$ is our desired concurrence point.