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:
$\frac{AM.DM}{2}.\frac{CN.FN}{2}=16 \implies AM.DM.CN.FN=64\\ \frac{S1}{S2} = \frac{AM.MD}{CN.FN}\\ \frac{S1}{\frac{MI.DM}{2}}=\frac{AM}{MI}\implies S1 = \frac{AM.DM}{2}\\ \frac{S2}{\frac{NI.FN}{2}}=\frac{CN}{NI}\implies S2 = \frac{CN.FN}{2}$
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On
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 $~MN = 4 \sqrt2$.
$$\angle A=\angle IDM = \angle DIB = \angle IBF = \angle FIN = \angle CFN$$
$$\frac{DM}{AM}=\frac{IM}{DM}=\frac{FI}{BF}=\frac{FN}{IN}=\frac{CN}{FN}=x$$
$$DM=x AM, IM=x DM = x^2 AM,$$ $$BF=ID=\sqrt{IM^2+DM^2}=x AM \sqrt{1+x^2}, FI=x BF=x^2 AM \sqrt{1+x^2}$$ $$FI^2=IN^2+FN^2=IN^2+(x IN)^2=(1+x^2) IN^2 \Rightarrow IN=\frac{FI}{\sqrt{1+x^2}}=x^2 AM$$ $$FN=x IN=x^3 AM, CN=x FN=x^4 AM$$
$$S_1=AM\cdot DM / 2 = x AM^2 / 2, S_2=FN\cdot CN / 2= x^7 AM^2 /2$$ $$S_1 S_2 = x^8 AM^4 / 4 \Rightarrow x^2 AM = \sqrt[4]{4S_1 S_2}$$ $$MN=IM+IN=2x^2 AM=\sqrt[4]{64 S_1 S_2}=4\sqrt{2} \rm{\ cm}$$