If the orthocenter of an isosceles triangle lies on the incircle of the triangle then

Let $ABC$ be the triangle in which $\mathrm{AB}=\mathrm{AC}.\text{ Let }\mathrm{I},\mathrm{H}$

respectively be the incentre and the orthocenter of the triangle. 

$\begin{array}{l}AI=r\mathrm{cosec}\frac{A}{2},AH=2R\cos A\text{ now }AI=AH+r\\ r\mathrm{cosec}\frac{A}{2}=2R\cos A+r\\ \Rightarrow 4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\left(\mathrm{cosec}\frac{A}{2}-1\right)\\ =2R\cos A\\ 2\sin \frac{B}{2}\sin \frac{C}{2}\left(1-\sin \frac{A}{2}\right)=\cos A\\ \Rightarrow 2{\sin }^2\frac{B}{2}\left(1-\sin \frac{A}{2}\right)=\cos A\\ \Rightarrow 2{\sin }^2\frac{B}{2}\left(1-\sin \left({90}^{\circ }-B\right)\right)=\cos \left({90}^{\circ }-2B\right)\end{array}$

$\begin{array}{l}\Rightarrow 2{\sin }^2\frac{B}{2}(1-\cos B)=\sin B-\cos 2B\\ (1-\cos B{)}^2=-\cos 2B\\ 1+{\cos }^{2}B-2\cos B=-\left(2{\cos }^{2}B-1\right)\\ \Rightarrow 3{\cos }^{2}B-2\cos B=0\\ \Rightarrow \cos B=\frac{2}{3}\\ \text{ OR use }H=r\end{array}$