shohvanilu wrote: In acute triangle $ABC$ $AD$ is bisector. $O$ is circumcenter, $H$ is orthocenter. If $AD=AC$ and $AC\perp OH$ . Find all of the value of $\angle ABC$ and $\angle ACB$. We have $AC\perp OH$ and $AC\perp BH$ $\Rightarrow B,O,H \in \overline{BH}$ $\Rightarrow \Delta ABC$ is the an isosceles triangle with $BA=BC$ Let $\widehat{ABC}=\alpha $ $\Rightarrow \widehat{BAC}=\widehat{BCA}=\frac{180^{\circ}-\alpha }{2}$ Triangle $ADC$ is isosceles with $AD=AC$ $\Rightarrow \widehat{ADC}=\widehat{DCA}=\frac{180^{\circ}-\alpha }{2}$ We have $\widehat{DAC}+\widehat{ACD}+\widehat{CDA}=180^{\circ}$ $\Rightarrow \frac{180^{\circ}-\alpha }{2}.2+\frac{180^{\circ}-\alpha }{4}=180^{\circ}$ $\Rightarrow \alpha=36^{\circ}$ $\Rightarrow \widehat{ABC}=36^{\circ}$ and $ \widehat{ACB}=72^{\circ}$

I think it's not original problem.In original version $ AD\perp OH $.

I think it's not original problem.In original version $ AD\perp OH $.

According to sardor's revised edition to this problem, whe have $ AD\perp OH\implies AH=R\implies\angle A=120^0, 60^0 $ rest is easy.