Apply Ceva's Theorem, I Don't Think

I will write a general solution for when the area of $\Delta AHE$ is $a$ , the area of $\Delta AHB$ is $b$ and the area of $\Delta BHD$ is $c$ .

Note that triangles on collinear bases with equal altitudes have their areas proportional to the lengths of their bases.

By this above property, we have that $\dfrac{EH}{HB} = \dfrac{a}{b}$ and $\dfrac {AH}{HD} = \dfrac {b}{c}$ . Draw line $CH$ and let the areas of $\Delta CHE$ and $\Delta CHD$ be $d$ and $e$ respectively. Notice that $\Delta CAH$ and $CHD$ have collinear bases and have equal altitude. Thus, $\dfrac {BH}{HE} = \dfrac {a+e}{d}$ . Similarly, $\dfrac {AH}{HD} = \dfrac {d}{c+e}$ . Thus, we have the following:

$\begin{aligned} \dfrac {b}{c} &= \dfrac {d}{c+e}\\ \dfrac {a}{b} &= \dfrac {a+e}{d} \end{aligned}$

Solving the simultaneous linear equations in terms of $d$ and $e$ , we get $d=\dfrac {ac(a+b)}{b^2-ac}$ and $e=\dfrac {ac(b+c)}{b^2-ac}$ , so the area of $CDHE=d+e=\dfrac {ac(a+2b+c)}{b^2-ac}$ . Substituting the values from the question is, we get the answer of 96. Thus, the answer is 96.

I will use the property that in the given quadrilateral $[∆AEH][∆BHD]=[∆EHD][∆AHB]$
Where [∆ABC] denotes Area of ∆ABC.
Plugging in the values, we get $[∆EHD]=\frac {8}{3}$ .
Note that triangles on collinear bases with equal altitudes have their areas proportional to the lengths of their bases, as stated by SHARKY sir. So, $\frac {CD}{DB}= \frac {[∆EDC]}{[∆EDB]} = \frac {[∆ACD]}{[∆ADB]}$ .
Again plugging in the values we get area of ∆ECD=93.33. So area of CEHD= 96