Let $\triangle ABC$ be a acute-angled triangle so that $AB>AC$ and $\angle BAC = 60^\circ$. Let $O$ be the circumcenter and $H$ the orthocenter. Let $OH$ intersect $AB$ and $AC$ in $P$ and $Q$ respectively. Show that $AQ=AP$. Or alternatively, show that $\angle AOP$ and/or $\angle AHQ$ equal $\angle ACB+90-\angle BAC$
I know $BAO=CAH=90-\angle ACB$, but that's all I've got. I need one of two above. An elementary solution is preferred.
On
The length of $AH$ in any $\triangle ABC$ is given by $2R\cos A$, and can be proven rather easily using the sine rule. Thus, in this case $\cos A=\frac12$, thus $AH =OH$. Hence, $\angle POA=\angle QHA$.
Of course, $\angle OAP = \angle OAB = 90^{\circ}-\angle C=\angle HAC = \angle HAQ$. Thus, by virtue of ASA criterion, $\triangle APO \cong \triangle AQH$.
Construction: CH is produced to cut the circumference at X. BX, when joined, cuts QP produced at M.
The pink angles are all $30^0$.
OM is the perpendicular bisector of BX. This makes the angles marked red are all equal to $30^0$.
The above is sufficient to say $\angle AQH = \angle APQ = 60^0$