There Are Too Many Things!

We will first find expressions for the areas.

$\begin{aligned} |ABMN| &= \dfrac{1}{2} \times AM \times BN \times \sin C\\ \dfrac{200}{\sin C} &= AM \times BN\\ |CMN| &= \dfrac{1}{2} \times CM \times CN \times \sin C\\ \dfrac{8}{\sin C} &= CM \times CN\\ |EFGH| &= \dfrac {1}{2} \times EG \times FH \times \sin C\\ \dfrac{2|EFGH|}{\sin C} &= EG \times FH \end{aligned}$

By Menelaus' in $\Delta ABC$ with line $DFG$ (no directed lengths),

$\dfrac {AD}{DB} \times \dfrac {BG}{GC} \times \dfrac {CF}{FA} = +1$

Because $AD$ and $AF$ are tangents to $\omega$ , $AD=AF$ . Thus,

$\dfrac {CF}{DB} \times \dfrac {BG}{GC} = +1$

$\dfrac {BG}{CG} = \dfrac {BD}{CF}$

Since $BD$ and $BE$ are tangents to $\omega$ , $BD=BE$ . Similarly, $CE=CF$ . Therefore,

$\dfrac {BG}{CG} = \dfrac {BE}{CE}$

$\dfrac {CG}{CE} = \dfrac {BG}{BE}$

By Componendo et Dividendo,

$\dfrac {CG+CE}{CG-CE} = \dfrac {BG+BE}{BG-BE}$

$\dfrac {EG}{EG-2CE} = \dfrac {EG+2BE}{EG}$

$\dfrac {2EN}{2EN-2EC} = \dfrac {2EN+2BE}{2EN}$

$\dfrac {EN}{CN} = \dfrac {BN}{EN}$

$EN^2 = BN \times CN$

Similarly, $FM^2 = CM \times AM$ . Now,

$EG^2 = 4 \times EN^2 = 4 \times BN \times CN$

$FH^2 = 4 \times FM^2 = 4 \times CM \times AM$

$(EG \times FH)^2 = 16 \times AM \times BN \times CM \times CN$

$\dfrac {4|EFGH|^2}{\sin^2 C} = 16 \times \dfrac {200}{\sin C} \times \dfrac {8}{\sin C}$

$|EFGH|^2 = 6400$

$|EFGH| = 80$

Therefore, the area of $EFGH$ is 80.

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For those who want the general form for the area of $EFGH$ , it is $4ab$ , where the area of $ABMN$ is $a^2$ and the area of $CMN$ is $b^2$ . Try to prove this!