# Triangle centers Orthocenter

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.

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