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

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    $\begingroup$ 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 $\endgroup$ – lioness99a Sep 25 '18 at 14:29

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$D$ 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 $A=150^\circ$.


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