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There is a point $O$ inside $\Delta ABC$. Join $A,O; B,O$ and $C,O$ and extend those lines. They will intersect $BC, AC$ and $AB$ at points $D, E$ and $F$ respectively. $AF:FB = 4:3$ and area of $\Delta BOF$ and $\Delta BOD$ is $60$ and $70$ square units respectively. Find the triangle with the largest area among $\Delta AOF, \Delta AOE, \Delta COE$ and $\Delta COD$ and write down the area of it.

Let $(XYZ)$ denote the area of triangle $XYZ$.
$\frac{4}{3}=\frac{AF}{BF}=\frac{(AOF)}{(BOF)}=\frac{(AOF)}{60}$
or $(AOF)=60$
Let (COD)=x.By Sheva's theorem $\frac{AF}{BF}\times \frac{BD}{CD}\times \frac{CE}{AE}=1$
or $\frac{4}{3}\times \frac{70}{x}\times \frac{CE}{AE}=1$
or $\frac{AE}{CE}=\frac{280}{3x}$
Again $\frac{AE}{CE}=\frac{(ABE)}{(CBE)}=\frac{(AOE)}{(COE)}=\frac{(AOB)}{(COB)}=\frac{140}{x+70}$
So $\frac{AE}{CE}=\frac{280}{3x}= \frac{140}{x+70}$
Solvig this equation we get $x=140$
So $(COD)=140$
After some calculation one will get $(AOE)=112$ and $(COE)=168$
So $(COD)$ has the largest area and and the area is 168.

We can use the Sheva's theorem to solve it.