Let line $CD$ intersect $(ABC)$ again at $E$. Since $\angle OHE=\angle OHB$, then $B$ and $E$ are symmetric wrt line $OH$, thus we get $HE=HB$. So $\angle HEB=\angle HBE$. Since $\angle HEB=\angle A$ and $\angle HBE=180^{\circ}-2\angle A$, we get $\angle A=60^{\circ}$. This value is obviously possible, for example when $ABC$ is equilateral.

Let $P$ be on $AC$, $Q$ on $AB$ such that $PQ$ is the angle bisector of $\angle{DHB}$. Let $BH$ meet $AC$ at $E$. Let $BE$ and $CD$ meet the circumcircle of $ABC$ at $M$ and $N$. Since $O$ lies on the bisector of $\angle{DHB}$, then $O$ is equidistant from lines $DH$ and $BH$, therefore $O$ is equidistant from chords $CN$ and $BM$ which implies they have the same length. Now, since points $B,C,M,N$ lie on the same circle and $BM=CN$ we must have $BC$ parallel to $MN$ or $CM$ parallel to $BN$. (1) If $MN$ is parallel to $BC$, $\angle{NCB}= \angle{MBC}$ which implies that $ABC$ is isosceles. Since $\angle{ACB}= 90-\frac{A}{2}$ and by an easy angle chase $\angle{APQ}=90-\frac{A}{2} $ then $PQ$ is paralel to $BC$. Then, since $O$ lies in $AH$, and $AH$ is perpendicular to $BC$, $AO$ is perpendicular to $PQ$ and therefore $O$ would lie on the external bisector of angle $\angle{DHB}$ which can't happen. (2) If $CM$ is parallel to $BN$, $\angle{MBN}= \angle{BMC}=\angle{BAC}$ Since $\angle{BAC}= A$ and by an easy angle chase we obtain $\angle{MBN}= \angle{MBA}+\angle{ABN}=90-A+90-A= 180-2A$ we get: $A= 180-2A $ $=>$ $A=60$

Let $X,Y$ be the projections of $O$ on CH and BH, $OX=OY$, we know $\frac{S_{COD}}{S_{BOE}}=\frac{h_c}{h_b}\frac{sin{(C+2A)}}{sin{(B+2C)}}=\frac{h_c}{h_b}$, where $E$ is the projection of $B$ on AC, so $\sin{(C+2A)}=\sin{(B+2C)}$, and both angles can't be greater than 90, so we have $2A=B+C$, so $A=60$. I used $2S_{XYZ}=XY*XZ*sin{\widehat{YXZ}}=YZ*h_X$