Question 147
In triangle $ABC$, the internal angle bisectors $BE$ and $CF$ intersect at point $I$. Let $F \in AB$, $E \in AC$. Let $D$ be the foot of the perpendicular from $I$ to $BC$. If $\angle A = 60^\circ$, prove that
$$
S_{BCF} + S_{BCE} = 3S_{BIC},
$$
where $S_{XYZ}$ denotes the area of triangle $XYZ$.
(Bilegdemberel Bat-Amgalan)
$$
S_{BCF} + S_{BCE} = 3S_{BIC},
$$
where $S_{XYZ}$ denotes the area of triangle $XYZ$.
(Bilegdemberel Bat-Amgalan)