We use complex numbers. WLOG, assume that the circumcircle of $\triangle ABC$ is the unit circle and that $a=1$. Then note that since $DM \perp AB$ where $m=\frac{1+b}{2}$, we must have \[\frac{2d-b-1}{2(b-1)}=-\frac{2/d-1/b-1}{2(1/b-1)} \implies d=\sqrt{b}\] Now it is well known that the reflection of a point $z$ over a chord $AB$ on the unit circle is $a+b-ab\bar z$. Thus $d'=1+b-b\sqrt{b}$. Similarly, $e'=1+c-c\sqrt{c}$. To make computations easier, we can let $b=b_1^2$ and $c=c_1^2$ so then $d'=1+b_1^2-b_1^3$ and $e'=1+c_1^2-c_1^3$. Now all that remains is to bash out the conditions for collinearity and concyclity which get slightly ugly, but are still very manageable in the time limit. I will edit in the full solution when I find time.

First by simple angle chasing we can observe that both $AD'HB$ and $AHE'C$ are con-cyclic $\angle D'HE'=\angle AHE'-\angle AHD'=(180-\angle ACE')-\angle ABD' =180 -\frac{(B+C)}{2}$ .If $A,D',E'$ are collinear, $\angle BAD'+\angle CAD'=A \Rightarrow \frac{(B+C)}{2}=A$ .From easy angle chase $D'OE= 180-A \Rightarrow D'HE'O$ are concyclic . If $H,D',E' ,O$ are cocncyclic.$\angle D'HE=\angle D'OE \Rightarrow \frac{(B+C)}{2}=A \Rightarrow \angle BAD'+\angle CAE'=A$ Done!

As $ AC>AB $ we arrange the following lines up to down $ BN,BE,CD,CM $ w.r.t $ O $ the circumcenter of $ \triangle ABC $ ,where $ M,N $ are intersections of circumcircle with $ CH,BH $ respectievly, due to this order clearly we have $ BE=CD\Leftrightarrow\angle BNE=\angle DMC $ and clearly we have $ HMDD' $ and $ HE'EN $ are isoscleses trapizoids, hence we have $ A,D',D $ are coliner$ \Leftrightarrow CE\parallel BD\Leftrightarrow BE=CD\Leftrightarrow \angle BNE=\angle DMC $ or $ \angle HE'O=\angle DD'H\Leftrightarrow $ $ D',H,E',O $ are concyclic, done!

NOTE: $A=\dfrac{B+C}{2}\Leftrightarrow 2.A=B+C\Leftrightarrow 3.A=A+B+C=180^0\Leftrightarrow \boxed{A=60^0}.$