Triangle Splits

As in the diagram, let $[EPC]=x$ and $[DPC]=y$ .

Note that triangles $BAP$ and $EAP$ have the same height (the perpendicular distance from $A$ to $BE$ ). Hence, $BP/PE$ is equal to the ratio of their areas, which is $\frac{20}{10}=2$ . So $BP=2PE$ .

Consider triangles $BCP$ and $ECP$ . We know that they have the same height, but their bases are in the ratio $2:1$ . Hence, their areas are also in the ratio $2:1$ . This gives us our first equation: $30+y=2x$ .

Now, note that triangles $DBP$ and $ABP$ have the same height, and their areas are in the ratio $3:2$ . Therefore, $DP=\frac{3}{2}PA$ .

Consider triangles $DCP$ and $ACP$ . They have the same height, but their bases are in the ratio $3:2$ . Then their areas are also in the ratio $3:2$ , which gives us our second equation: $y=\frac{3}{2}(10+x)$ .

Solving the two equations simultaneously gives $x=90$ and $y=150$ , so the area of triangle $ABC$ is $10+20+30+90+150=\boxed{300}.$