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P is a point in triangle $ABC$. The lines $AP$,$BP$, and $CP$ intersect the sides $BC$,$CA$, and $AB$ at points $D$,$E$, and $F$, respectively.
If $[BDP]=10$, $[DPC]=16$, $[APB]=210$, what is $[APC]$?
The line segments $BD$ and $DC$ are in the ratio $10$ to $16$, because with respect to these bases $\triangle BDP$ and $\triangle DPC$ have the same height.
So the areas of $\triangle ADB$ and $\triangle ADC$ are in the same ratio (bases $BD$ and $DC$, same height) and therefore by subtraction so are the areas of $\triangle APB$ and $\triangle APC$.
It follows that $\triangle APC$ has area $210\cdot \frac{16}{10}$.