An acute $\triangle ABC$, inscribed in a circle $k$ with radii $R$, is given. Point $H$ is the orthocenter of $\triangle ABC$ and $AH = R$. Find $\angle BAC$. (Answer: $60^\circ$)
$AD$ $-$ diameter, thus $\angle ACD = \angle ABD = 90^\circ$. Also $HBDC$ is parallelogram because ($HC || BD$, $HB || CD$). It seems useless and I don't know how to continue.