<s>[INST] Find the area of ​the shaded region. Given, $MT=2$ and $AC=8$. Calculate the area of ​​the shaded region.

$\triangle BMT_(notable)\implies (a, 2a, a\sqrt5)\\
\therefore MT = 2, BT=4\\
\triangle ABF \sim \triangle MBT \implies k = \frac{MT}{BT }=\frac{1}{2} =\frac{AF}{BF}\therefore AF =2.2 = 4, BF = 2.4 = 8 \\
S\triangle AMC = S\triangle MCB$
Missing a detail,,,
I managed to solve
 [/INST] 
$MT=2, BT = 4, BM = 2\sqrt5\\
\triangle AGC \sim \triangle MTB\\
\frac{GC}{TB}=\frac{AC}{MB}\\
\frac{GC}{4}=\frac{8}{2\sqrt5}\\
∴GC=\frac{16}{\sqrt5}\\
S_{\triangle MBC} =\frac{1}{2}.MB⋅GC=\frac{1}{2}\cdot 2\sqrt5 \cdot \frac{16}{\sqrt5}=16$
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