Given a convex pentagon $ABCDE$, with $\angle BAC = \angle ABE = \angle DEA - 90^o$, $\angle BCA = \angle ADE$ and also $BC = ED$. Prove that $BCDE$ is parallelogram.
Denote $\angle ABE = \alpha$ , $\angle BCA = \beta$ and $P$ be intersection point of $AC$ with line parallel to $AB$ from $E$.
Note that, $EP \parallel AB$ and $\angle PAB = \angle CAB = \angle ABE = \alpha \Longrightarrow \angle APE = \alpha$.
$\angle APE = \alpha= \angle EBA \Longrightarrow BAEP$ is isosceles trapezoid with $BP = AE$.
Denote $\angle DEP = \theta$.
$\angle DEA = 90^0 + \alpha \Longrightarrow \angle BAE = 90^0 - \theta$.
Now by sinus lawes in $\triangle BAE$ and $\triangle AED$ we have:
$$\frac{\sin{(90^0-\theta)}}{\sin{\alpha}} = \frac{AB}{AE} = \frac{\frac{BC \cdot \sin{\beta}}{\sin{\alpha}}}{\frac{DE \cdot \sin{\beta}}{\sin{(90^0-\alpha-\beta)}}} = \frac{\sin{(90^0-\alpha-\beta)}}{\sin{\alpha}} \Longrightarrow \theta = \alpha + \beta $$Now we have $\angle CBE + \angle DEB = (180^0-2\alpha -\beta) + 2\alpha + \beta = 180^0 \Longrightarrow BC \parallel DE$.
So we have $BC \parallel DE$ and $BC = DE$. Hence $BCDE$ is parallelogram
$Q.E.D. \blacksquare$