BdMO Online Forum

It is easy to see that $\triangle ABO_{B} \sim \triangle ACO_{C}$.
So,$A$ is the center of spiral similarity which takes $BO_{B}$ to $CO_{C}$.Then from a well known result,we get that $A$ is the Miquel Point of cyclic quadrilateral $BCO_{B}O_{C}$.

Now let $BC \cap O_{B}O_{C}=P$,$AH \cap BC=Q$ and $AD \cap O_{B}O_{C}=R$.Then $XA \perp AP$,$AD \perp O_{B}O_{C}$ and $AQ \perp BC$.Therefore,$A,R,Q,P$ are cyclic.So,

$\measuredangle DAH=\measuredangle EAQ=\measuredangle EPQ=\measuredangle APE$ (Since $PBO_{B}A$ is cyclic and $O_{B}A=O_{B}B$) $=\dfrac {\pi} {2}-\measuredangle EAP=\measuredangle XAE=\measuredangle XAD$.

Q.E.D