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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.

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Hint: Let $P$ be the intersection of ray $HE$ with the circumcircle of $\triangle ABC$. Then, $\angle PAC = \angle PBC = \angle FAC = 90^\circ - C$.

It follows that $\triangle PAH$ is isosceles at $A$, and that $AP = AH = R$. Consequently $\triangle APO$ is an equilateral triangle.

Finally, $$\angle PAO = (90^\circ - C) + \angle CAO = \angle OAB + \angle CAO = A,$$ and we are done.

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  • $\begingroup$ Why are the angles $OAB=PAC=90-C$ $\endgroup$ – Fareed AF May 7 at 5:36
  • $\begingroup$ $\angle OAB = 90 - \angle ADB = 90 - \angle C$. $\endgroup$ – Quang Hoang May 7 at 12:35

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