Let $ABCD$ be a right-angled trapezoid and $M$ be the midpoint of its greater lateral side $CD$. Circumcircles $\omega_{1}$ and $\omega_{2}$ of triangles $BCM$ and $AMD$ meet for the second time at point $E$. Let $ED$ meet $\omega_{1}$ at point $F$, and $FB$ meet $AD$ at point $G$. Prove that $GM$ bisects angle $BGD$.
Problem
Source: Sharygin Finals 2022 8.2
Tags: geometry, trapezoid
g0USinsane777
02.11.2022 13:22
Note that, $\angle BEM = 180 - \angle C = \angle D $, which means that $E$ is a common point to cyclic quadrilaterals $EBCM$ and $AEMD$ and $E$ lies on line $\overline {AB}$. $\angle EMC = 90^\circ$ and $\angle EFC = \angle EMC = 90^\circ$ as quad $EFMC$ is cyclic. Thus, $\angle CFD = 90 ^\circ$ which implies $FM = CM = MD$. Now, let $\angle ABG = \alpha$, then $\angle BGA = 90 - \alpha$ and as $BEMF$ is cyclic, $\angle EMF = \alpha$. Which means that, $\angle BGA = \angle FMD = 90 - \alpha$. This implies that quad $FMDG$ is cyclic. And as $FM=MD$, we get $\angle MGF = \angle MGD$. Hence, $GM$ bisects $BGD$.
InterLoop
21.07.2024 17:35
First of all note that $EM \perp CD$. We claim $DMGF$ concyclic, since $$\measuredangle MDG = \measuredangle CDA =\measuredangle DCB = \measuredangle MCB = \measuredangle MFB = \measuredangle MFG$$Clearly $MD = MF$ is equivalent to $\angle DFC = 90$, which is true as $\angle DFC = \angle EFC = \angle EBC = 90$.