A BMO problem

[Phlembac Adib Hasan]

Points $A,B,C,D$ and $E$ lie, in that order, on a circle and the lines $AB$ and $ED$ are parallel. Prove that $\angle ABC = 90^{\circ} \Longleftrightarrow AC^2 = BD^2 + CE^2$.

[SANZEED]

Here, $\angle ABC=90^{\circ}$. Since $ABCE$ is a cyclic quadrilateral, $\angle AEC=90^{\circ}$.
So in $\triangle AEC, AC^2=AE^2+EC^2$.
But in the cyclic quadrilateral $ABDE, AB\parallel ED$, that is, $ABDE$ is a cyclic trapizoid,and thus it's isosceles & $AE=BD$.
Combining this results, $AC^2=BD^2+CE^2$
To prove that $\angle ABC=90^{\circ}$, note again that $ABDE$ is a cyclic trapizoid,thus isosceles, thus $AE=BD$.
Now $AC^2=BD^2+CE^2=AE^2+CE^2$, so by the converse of Pythagoras, $\angle AEC=90^{\circ}$ and so is $\angle ABC$. Proved!