$A$ square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

(N/A) $ABC$ is an isosceles right triangle with $AB = BC$. $A$ square $BFED$ is inscribed in triangle $ABC$ such that $\angle B$ is common and $\angle B = 90^{\circ}$.
In $\triangle ADE$ and $\triangle EFC$,we have:
$DE = EF$ (Sides of a square are equal) ... $(1)$
$\angle 1 + \angle 2 = 180^{\circ}$ (Linear pair axiom)
Since $\angle 1 = 90^{\circ}$ (angle of a square),$\angle 2 = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
Similarly,$\angle 4 = 90^{\circ}$.
Therefore,$\angle 2 = \angle 4 = 90^{\circ}$ ... $(2)$
Since $AB = BC$ (given),$\angle A = \angle C$ ... $(3)$ (Angles opposite to equal sides are equal).
From $(1)$,$(2)$,and $(3)$,by $AAS$ congruence rule,$\triangle ADE \cong \triangle EFC$.
Hence,$AE = EC$ by $CPCT$,which means $E$ bisects the hypotenuse $AC$.