# Incenter geometry problem

Let $$I$$ be the incenter of $$\triangle ABC$$ and let $$\overline{AI}$$ meet the circumcircle of $$\triangle ABC$$ at $$D$$. Denote the feet of perpendicular from $$I$$ to $$\overline{BD}$$ and $$\overline{CD}$$ by $$E$$ and $$F$$ respectively. If $$\overline{IE} + \overline{IF} = \frac{\overline{AD}}2$$ calculate $$\angle BAC$$

I have used a few formulae related to triangles such as $$\overline{AI} = r\cdot\csc\bigg(\frac A2\bigg)$$ but I am unable to deduce anything

• Have you tried drawing a diagram? If so, can you add it here so we can see what you have done and maybe help if you have got something wrong – lioness99a Sep 25 '18 at 14:29

$$D$$ is the midpoint of the minor $$BC$$-arc in the circumcircle of $$ABC$$ and $$IE+IF = DI\,(\sin B+\sin C).$$ On the other hand $$DI=DB=DC$$, hence $$\frac{AD}{DI}=\frac{AD}{DB}=\frac{\sin\widehat{ABD}}{\sin\widehat{BAD}}=\frac{\sin(B+A/2)}{\sin(A/2)}$$ and the constraint $$IE+IF = \frac{1}{2} AD$$ is equivalent to the constraint $$\sin B+\sin C = \frac{\sin(B+A/2)}{2\sin(A/2)}$$ or to the constraint $$2\sin\frac{B+C}{2} = \frac{1}{2\sin(A/2)}$$ or to the constraint $$2\cos\frac{A}{2}\sin\frac{A}{2} = \frac{1}{2}$$ from which it follows that $$\sin A=\frac{1}{2}$$ and $$\color{red}{A=30^\circ}$$ or $$A=150^\circ$$.