geometrytriangles
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
Here is a possible picture for the already solved problem:
Well, how could you fill in the details without a faithful picture?! (And what is the "reflection of $N$ across $S$" by construction, the point $N$ is reflected with respect to the center $S$ or conversely?!)
COMMENT:
You correctly found that $\triangle IDE≈\triangle AFM$.We rewrite the required relation as:
$\frac{AF}{AM}=\frac{ID}{IE}$
That is triangles IAE and DIE must be similar. These two triangles have common angle $\angle IED$, therefore we must have:
$\angle IDE=\angle EIA$
This is not possible because:
$\angle IDE=\angle BDE (=90^o)+\angle BDI$
$\angle EIA=\angle EIF (≠90^o)+\angle FIA(=BDI)$
Best Answer
$D$ 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 $A=150^\circ$.