Pure Congruency and Similarity.

It's IMO G4. I am posting my reverse solution on AoPS. Let $J$ be the reflection of $I$ in the line $AC$ and $AI \cap BC = P.$ In $\Delta BIP,$ we have: $\dfrac{IB}{IP} = \dfrac{sin90^{\circ}}{sin \angle IBP}$

$\dfrac{IB}{2 \times IP} = \dfrac{1}{2 \times sin \angle 45 - A/4}$

$\dfrac{IB}{IJ} = \dfrac{1}{2 \times sin[ \angle 45 - A/4]} --1$

In $\Delta EDI,$ we have:: $\dfrac{DE}{DI} = \dfrac{DE}{DJ} = \dfrac{sin 45 + A/4}{sin 90+A/2} = \dfrac{sin 45 + A/4}{sin 2 \times 45 + A/4} = \dfrac{sin 45 + A/4}{2 \times [sin 45 + A/4][cos 45 +A/4]} = \dfrac{1}{2 \times [sin 45 - A/4]}$ $--2$

Now, angle chasing is telling us that $\angle BIJ = EDJ$ Therefore, from $[1] , [2]$ we get $\Delta BIJ \sim \Delta EDI => \angle JBI = \angle JBD = \angle JED => BEDI$ is cyclic.

We can also backup the whole trig part through some constructions like producing side $IP$ to increase its length by a factor of $2$ and also reflecting $DE$ to a point on $AP$ such that $DE' = DE$ . Now prove the two new triangles similar.

Note : The statement of the problem asks for converse.