In triangle ABC, E is a point on AC and F is a point on AB. The lines BE and CF intersect at D. If the areas of triangles BDF, BCD nad CDE are 3,7 and 7 respectively, what is the area of the quadrilateral AEDF
$\displaystyle area(BDC)=area(CDE)\Rightarrow BD=DE\Rightarrow area(ABD)=area(ADE)$
Let $\displaystyle area(ADE)=x, \ area(ADF)=y$. Then, $\displaystyle x=y+3$
$\displaystyle \frac{area(BDF)}{area(BDC)}=\frac{FD}{DC}=\frac{3} {7}=\frac{area(AFD)}{area(ADC)}$
Then $\displaystyle \frac{y}{x+7}=\frac{3}{7}\Rightarrow 7y=3x+21$
Solving the system $\displaystyle \left\{\begin{array}{ll}x=y+3\\7y=3x+21\end{array} \right.$ we get $\displaystyle x=\frac{21}{2}, \ y=\frac{15}{2}$
$\displaystyle area(AEDF)=x+y=18$