In a triangle $ABC$ line joining circumcenter and orthocenter is paralel to line $BC$. Find the range of $\angle A$
Solution i try:
From figure $HD =2R\cos B\cos C= OE$ and $\displaystyle \angle COE =A$
$ 2R\cos B\cos C=R\cos A$ here $R$ is circumradius of circle
$2\cos B \cos C =\cos A$ how to get range of $A$ from that point?
As $C=\pi-A-B$, you have: $\cos C=-\cos(A+B)=\cos A\cos B-\sin A\sin B$. Substituting that into your equation $-2\cos B\cos C+\cos A=0$ gives: $$ \cos A(1+\cos^2B)-2\sin A\sin B\cos B=0, $$ that is: $$ \cos A(2+\cos 2B)-\sin A\sin 2B=0, $$ or: $$ \tan A ={2+\cos 2B\over \sin 2B}. $$ Notice that we can choose $B$ such that $B\le C$ and thus $B\le\pi/2$. If you now consider the function $$f(x)={2+\cos 2x\over \sin 2x}$$ for $x\in(0,\pi/2)$, you can easily find its minimum to be $\sqrt3$, hence $\tan A\ge\sqrt3$ and $60°\le A\le 90°$.