Post Number:#2 by aritra barua » Mon May 29, 2017 12:34 am
Let the reflection of $DM$ over $M$ be $D'M$.Thus we concurr that $AD'BD$ is a parallelogram as $M$ is designed to be the midpoint of $AB$.Let the midpoint of $AC$ be $X$.Thus we have $AX$=$CX$; $ZX$ as a common side of triangles $ZXA$ and $ZXC$;$\angle ZXA$=$\angle ZXC$.So,by $SAS$,we have $\bigtriangleup ZXA$ $\cong$ $\bigtriangleup ZXC$ which follows $ZA$=$ZC$.Now let $\angle BAD$ be $\alpha$ and $\angle D'AB$ to be $\gamma $.Now, trivial angle chasing shows that $\angle D'AZ=270^{\circ}-\alpha$-$\gamma$.Again $\angle DCZ$=$\angle BCD$+$\angle ACB$+$\angle ACZ=90^{\circ}+\angle ACB$.Since $\angle ACB$=$\angle ADB$=$\angle AD'B$,it's clear that $\angle DCZ$=$\angle D'AZ$=270-$\alpha$-$\gamma$.So,by applying $SAS$ again,we get $\bigtriangleup D'AZ$ $\cong$ $\bigtriangleup DCZ$,hence $DZ$=$D'Z$.Therefore, $\bigtriangleup DD'Z$ is isosceles showing that $ZM$ $\perp$ $DD'$.So,$\angle ZMD=90^{\circ}$,which was our intended goal leaving $AMDZ$ cyclic.So,we are done.