Let $M$ and $N$ be the midpoints of $AB$ and $AC$.$MN$ meets $DC$ at $X$.Then $\angle AMX=\angle ABC=\angle ADX$.So $A$,$M$,$D$,$X$ are cyclic.Then $\angle NXC=\angle MXD=\angle MAD=\angle DAC=\angle AZN=\angle CZN$ implies $Z$,$N$,$C$,$X$ are cyclic so $\angle CXZ=\angle DXZ=90$,$\angle DAZ=90$. So...