<s>[INST] What's the area of the shaded regions in the triangle below? For reference: In triangle ABC, $S_1$ and $S_2$ are areas of the shaded regions. If $S_1 \cdot{S}_2=16 cm^4$, calculate $MN$.

My progress:
$\frac{AM.DM}{2}.\frac{CN.FN}{2}=16 \implies AM.DM.CN.FN=64\\
\frac{S1}{S2} = \frac{AM.MD}{CN.FN}\\
\frac{S1}{\frac{MI.DM}{2}}=\frac{AM}{MI}\implies S1 = \frac{AM.DM}{2}\\
\frac{S2}{\frac{NI.FN}{2}}=\frac{CN}{NI}\implies S2 = \frac{CN.FN}{2}$
.....????
 [/INST] Making a few observations does reduce the work considerably. Say $DM = x, FN = y$

First, observe that $~\displaystyle \frac{x}{AM} = \frac{CN}{y} \implies AM \cdot CN = xy$
$s_1 \cdot s_2 = \frac 12 AM \cdot x \cdot \frac 12 CN \cdot y \implies xy = 2 \sqrt{s_1 s_2} = 8$
Next observation: Given $BDIF$ is a rectangle, diagonals $BI$ and $FD$ are equal and $O$ is the midpoint of both diagonals. We also have, $DM \parallel OI \parallel FN$
That leads to $~OI = \dfrac{x+y}{2} \implies FD = BI = x + y$
Finally using Pythagoras, $MN^2 = (x+y)^2 - (y-x)^2 = 4xy = 32~$ and $~MN = 4 \sqrt2$.
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