Extent $AM$ to meet $BC$ at $L$ and $BD$ at $O$. Now ∆$CML$~∆$MAD$ let $BC=X$ and$CM=2y,MD=3y$.$MD/CM$=$AM/ML$=AD/CL$ so $CL=2x$. Now $BL=3x=AD$ and $BL||AD$ so $BLAD$ is a $|| gm. And now ∆$AOB$ is congruent to ∆$OLD$so$AB=LD=3x$ so $ABLD$ is a rhombus and the diagonals of the rhombus bisects each other at $90$° .