Easy, ruins

Let $y$ be the area of $\triangle AEO$ and $a$ be the area of $\triangle COD$ . Recall that the areas of triangles with equal altitudes are proportional to the bases of the triangles. We have

$\dfrac{DC}{DB}=\dfrac{A_{ADC}}{A_{ADB}}=\dfrac{A_{ODC}}{A_{ODB}}$

$\dfrac{84+a+y}{40+30+35}=\dfrac{a}{35}$ $\implies$ $35(84+a+y)=105a$ $\implies$ $2940+35y=70a$ $\color{#D61F06}(1)$

$\dfrac{AF}{FB}=\dfrac{A_{CFA}}{A_{CFB}}=\dfrac{A_{OFA}}{A_{OFB}}$

$\dfrac{84+y+40}{30+35+a}=\dfrac{40}{30}$ $\implies$ $30(124+y)=40(65+a)$ $\implies$ $1120+30y=40a$ $\color{#D61F06}(2)$

Solve for $a$ in terms of $y$ in $\color{#D61F06}(2)$ then substitute in $\color{#D61F06}(1)$ , we have

$a=\dfrac{1120+30y}{40}$

$2940+35y=70\left(\dfrac{1120+30y}{40}\right)=1.75(1120+30y)=1960+52.5y$

$2940-1960=52.5y-35y$

$980=17.5y$

$\boxed{y=56}$