Confusing centres

$SOLUTION:$

We start by proving $\triangle ABC$ is isosceles. Then we proceed by chasing angles and prove $BDPH$ is a cyclic quad.

Let $\angle B$ be 2x. And $\angle C$ be 2y. Then $\angle A$ will be 180 - 2x - 2y.

As $AO$ bisects $\angle A$ , $\angle OAE$ = 90 - x - y.

$O$ is the circumcentre of $\triangle ABC$ . So $\angle OAE$ = 90 - 2x.

Equating , we get 90 - 2x = 90 - x - y $\Rightarrow$ x = y.

Thus $\triangle ABC$ is isosceles.

Let $OB$ meet $DE$ at $P$ . The midpoint of $OH$ be $N$ . Let $M$ be the midpoint of $BC$

As O is the incentre of $\triangle ADE$ , $\angle ADO$ = $\angle ODN$ = x.

As $ON$ = $NH$ and $DN$ = $NE$ and $OH$ $\perp$ $DE$ , ODHE is a Rhombus. $\Rightarrow$ $\angle NDH$ = x.

$\angle BDH$ = 180 - 3x. and $\angle BDN$ = 180 - 2x.

$\angle BOH$ = $\angle BAO$ + $\angle OBA$ = 90 - 2x + 90 - 2x = 180 - 4x. = $\angle NHP$ .

$\angle BHM$ = $\frac {180 - \angle A}{2}$ = $\frac {4x}{2}$ = 2x.

$\angle BHN$ = 180 - 2x.

$\angle BHP$ = $\angle BHN$ - $\angle NHP$ = 180 - 2x - ( 180 - 4x ) = 2x.

Thus we have $\angle BDP$ = $180 - \angle BHP$

So BDPH is a cyclic quadrilateral.

So $\angle BDH$ = $\angle BPH$

Also $\angle BPH$ = $\angle POH$ + $\angle OHP$ = 180 - 4x + 180 - 4x = 360 - 8x.

So 180 - 3x = 360 - 8x $\Rightarrow$ x = 36.

$\angle A$ = 180 - 4x = 180 - 4*36 = $\boxed {36 }$