6. The area of a triangle $A B C$ is $40 \mathrm{~cm}^{2}$. Points $D, E$ and $F$ are on sides $A B, B C$ and $C A$ respectively, as shown in the figure below. If $A D=3 \mathrm{~cm}, D B=5 \mathrm{~cm}$, and the area of triangle $A B E$ is equal to the area of quadrilateral $D B E F$, find the area of triangle $A E C$ in $\mathrm{cm}^{2}$.
(A) 11
(B) 12
(C) 13
(D) 14
(E) 15
6. Answer: (E)
Since area of triangle $A B E$ is equal to area of quadrilateral $D B E F$, we see that area of $\triangle D E A=$ area of $\triangle D E F$. This implies that $D E$ is parallel to $A F$.
Thus $\frac{C E}{C B}=\frac{A D}{A B}=\frac{3}{8}$. Since $\frac{\text { areaof } \triangle A E C}{\text { area of } \triangle A B C}=\frac{C E}{C B}$, it follows that area of $\triangle A E C=\frac{3}{8} \times 40=15 \mathrm{~cm}^{2}$.