Partitioning a triangle

Recall that the areas of triangles with equal altitudes are proportional to the bases of the triangles. Considering the diagram, we have $\dfrac{AF}{FB}=\dfrac{A_{CFA}}{A_{CFB}}=\dfrac{A_{EFA}}{A_{EFB}}$

$\dfrac{a+b+150}{50+150}=\dfrac{a}{50}$

$\dfrac{a+b+150}{200}=\dfrac{a}{50}$

$\dfrac{a+b+150}{4}=a$

$b=3a-150$ $\implies$ $\color{plum}\boxed{1}$

$\dfrac{AD}{DC}=\dfrac{A_{BDA}}{A_{BDC}}=\dfrac{A_{EDA}}{A_{EDC}}$

$\dfrac{a+b+50}{150+150}=\dfrac{b}{150}$

$\dfrac{a+b+50}{300}=\dfrac{b}{150}$

$\dfrac{a+b+50}{2}=b$

$a+b+50=2b$

$a-b=-50$ $\implies$ $\color{plum}\boxed{2}$

Substitute $\color{plum}\boxed{1}$ in $\color{plum}\boxed{2}$ .

$a-(3a-150)=-50$

$a-3a+150=-50$

$-2a=-200$

$a=100$

It follows that

$b=3(100)-150=150$

The desired answer is $a+b=100+150=$ $\boxed{250}$ .