# Question about area and triangle

Problem:

Consider the following diagram. in $\triangle$ABC:

Areas:

$\triangle$AOM = a

$\triangle$POC = b

$\triangle$NOC = c

$\triangle$BON = d.

Find the area of $\triangle$MOB and $\triangle$AOP in terms of a,b,c,d. Note that AN, BP and CM are not necessarily medians and $\triangle$ABC is not a special triangle.

diagram:

Consider the triangle $BOA$ and $OPA$, they share the same height, hence the ratio of their areas is the same as the ratio of their bases: $$\frac{a+y}{x}=\frac{BO}{OP}.$$ The same can be said of the triangles $BOC$ and $POC$: $$\frac{d+c}{b}=\frac{BO}{OP}.$$ By the same argument $$\frac{x+b}{a}=\frac{OC}{MO}=\frac{c+d}{y}.$$ It remains only to solve a system of two equations.
To help to solve the equation we can use the fact that $$adb=xyc$$ which you can check writing each area piece with the formula $1/2l_1l_2 \sin u.$