I'm assuming $O$ is the center of the circle?

lpieleanu wrote: I'm assuming $O$ is the center of the circle? a_507_bc wrote: Fix circle with center $O$, diameter $AB$ it says O is the center

I edited it after his comment.

Come on! This is an easy complex bash.

Note that $BOED$ is cyclic iff $\angle BOD=90^{\circ}$. Now $CE\cdot ED=OA^2-OE^2$ and this is maximal iff $OE$ is minimal. As $D$ moves along arc $AB$, $E$ moves along a semicircle with radius $CB$ so $OE$ is minimal iff $E$ is the midpoint of arc $CB$ (of the semicircle). Finally $45^{\circ}=\angle ECB= \angle DCB= \angle DAB$ So $D$ is the midpoint of arc $AB$ equivalently $\angle DOB=90^{\circ}$.

Attachments:

Note $B$ spirals $CE$ to $AD$ and $CA$ to $ED$. Hence $\frac{ED}{CA}=\frac{BD}{AB}$ and $\frac{EC}{AD}=\frac{BC}{AB}$. This means $$ED\times EC=\frac{BD\times BC\times CA\times AD}{AB^2}.$$The only thing that matters is $BD\times AD$, to maximise $D$ is the midpoint of arc.

Let $a=-1,b=1$, wlog $C$ is on upper side and $D$ on the lower side in the argand plane. $e=\tfrac{1}{2}(1+c+d-cd)$ \[|c-e||d-e| = \frac 14|1-cd -c + d||1-cd + c-d| = \frac 14 |(1+d)(1-c)||(1+c)(1-d)| = \frac14|1-c^2||1-d^2|\]for a fixed $c$ this expression is maximized when $|1 - d^2| = 1 + |d^2| = 2$ (by triangle inequality) which occurs at $d=-i$ i.e. $DO\perp AB \implies BOED$ is cyclic.