Triangles and Semicircles. [duplicate]
Let the orthocentre of the original triangle be $H$. Observe that, $AE\cdot EC=HE\cdot BE$ since $\triangle AEB\sim \triangle HEC$.
Thus, $EY^2=AE\cdot EC=HE\cdot BE$
Hence, $[\triangle ACY]^2=\frac {1}{4} AC^2\cdot HE\cdot BE=\left(\frac{1}{2} AC\cdot HE\right)\left(\frac {1}{2}AC\cdot BE\right)=[\triangle AHC]\cdot [\triangle ABC]$
Similarly,
$[\triangle BXC]^2=[\triangle BHC]\cdot [\triangle ABC]$
$[\triangle ABZ]^2=[\triangle AHB]\cdot [\triangle ABC]$
Adding these up gives the desired result.