Dear Mathlinkers, any ideas? Sincerely Jean-Louis

Let $I$ be the incenter of $\triangle ABC$ and $r$ denote its inradius. Then $\frac{S_{\triangle BMC}}{S_{\triangle CND}} = \frac{MK \cdot BD}{NL \cdot CD}$ It suffices to show that $\frac{MK \cdot BD}{DK} = \frac{NL \cdot CD}{DL}$. We have $MK \cdot FK = DK^2$ so $\frac{MK \cdot BD}{DK} = \frac{DK \cdot BD}{FK} = \frac{DK \cdot BF}{FK}$. Let $J$ be the midpoint of segment $FM$. Since $\angle IJM = 90^{\circ}$, we see $IJKD$ is a rectangle. Note that $\angle IFJ = 90^{\circ} - \angle BFK = \angle FBK$, so $\triangle IFJ \sim \triangle FBK$. Now $\frac{DK \cdot BF}{FK} = \frac{DK \cdot IF}{IJ} = IF = r$. Similarily, $\frac{NL \cdot CD}{DL} = r$. Hence $\frac{MK \cdot BD}{DK} = r = \frac{NL \cdot CD}{DL}$ and we are done.

What does $S$ denote? Area or semiperimeter? @brlow thanks, but I could not understand the first step of the previous solution itself

big $S$ means area edit: the tag has also areas, also reading the 1st solution you may understand that it refers to areas

IGO 2014 q2 wrote: The inscribed circle of $\triangle ABC$ touches $BC, AC$ and $AB$ at $D,E$ and $F$ respectively. Denote the perpendicular foots from $F, E$ to $BC$ by $K, L$ respectively. Let the second intersection of these perpendiculars with the incircle be $M, N$ respectively. Show that $\frac{{{S}_{\triangle BMD}}}{{{S}_{\triangle CND}}}=\frac{DK}{DL}$ Let angle $C = \gamma, B = \beta, A=\alpha$, so \[\frac{{{S}_{\triangle BMD}}}{{{S}_{\triangle CND}}}=\frac{BD \cdot MK}{NL\cdot CD} = \frac{KD\cdot BD\cdot \cot(\beta/2)}{\cot(\gamma/2)\cdot CD}=\frac{KD}{DL}\]since \[\frac{\cot(\beta/2)}{\cot(\gamma/2)}=\frac{s-c}{s-b}\]by law of cotangents.