Let $ABC$ be an acute triangle with $AB<AC$. The angle bisector of $BAC$ intersects the side $BC$ and the circumcircle of $ABC$ at $D$ and $M\neq A$, respectively. Points $X$ and $Y$ are chosen so that $MX \perp AB$, $BX \perp MB$, $MY \perp AC$, and $CY \perp MC$. Prove that the points $X,D,Y$ are collinear.
Problem
Source: Poland 73-3-1
Tags: geometry
timon92
01.04.2022 02:18
This problem was proposed by Burii.
Tintarn
01.04.2022 11:17
Let us prove that $\angle MXD=\angle MXY$.
Indeed, $\angle DMX=90^\circ-\frac{\alpha}{2}=90^\circ-\angle MBC=\angle DBX$ so that $BMDX$ is cyclic.
Hence $\angle MXD=\angle MBD=\frac{\alpha}{2}$.
Moreover, let $N$ be the point diametrically opposite to $M$ so that $B,X,N$ and $C,Y,N$ are collinear. Now $\angle YMX=\angle AMX+\angle YMA=2 \cdot \left(90^\circ-\frac{\alpha}{2}\right)=180^\circ-\alpha=180^\circ-\angle XNY$ so that $XMYN$ is cyclic.
Hence $\angle MXY=\angle MNY=\angle MNC=\frac{\alpha}{2}$ as desired.
PROF65
01.04.2022 13:17
Let $N$ the midarc of $AB$ containing $A$; since $BM\perp BX$ then $B,X,N$ are colinear idem $N,Y,C$ are colinear . $\angle XMY=\angle (\perp AB,\perp AC)=\angle BAC$ implies $MNXY$ is cyclic and $MX=MY$ more $\angle YMA=\angle (\perp AC,\perp AN)=\angle CAN =\angle CMN, \angle AMX=\angle (\perp AN,\perp AB)=\angle NAB=\angle NMB$ thus $MA $ is the bisector of $\angle XMY$ let $D'=MA\cap XY$ it suffices to show that $D=D'$ indeed since $MXY$ isoceles then $D'$ is the projection of$M$ on $XY$ besides $B,C,$ are the feet of $M$ on $NX,NY$ then by simson line $B,C,D'$ are collinear therefore the result follows . My regards RH HAS
Mogmog8
05.04.2022 21:49
Claim: $BDMX$ is cyclic. Proof. Notice \begin{align*}\measuredangle DMX&=\measuredangle AMB+\measuredangle BMX\\&=\measuredangle ACB+\measuredangle XBE\\&=\measuredangle ACB+\measuredangle NCA\\&=\measuredangle NAB\\&=\measuredangle MAB+90\\&=\measuredangle CBM+90\\&=\measuredangle DBX.\end{align*}$\blacksquare$ Similarly, $CMDY$ is cyclic. Hence, $\measuredangle XDM=90=\measuredangle YDM.$ $\square$