32. Given that $A B C D$ is a square. Points $E$ and $F$ lie on the side $B C$ and $C D$ respectively. such that $B E=C F=\frac{1}{3} A B$. $G$ is the intersection of $B F$ and $D E$. If
$$
\frac{\text { Area of } A B G D}{\text { Area of } A B C D}=\frac{m}{n}
$$
is in its lowest term, find the value of $m+n$.
32. Ans: 23.
Join $B D$ and $C G$ and note that $\frac{D F}{F C}=2$.

Assume the length of $A B$ is 1 . Let the area of $\triangle B G E$ and $\triangle F G C$ be $x$ and $y$ respectively. Then the areas of $\triangle E G C$ and $\triangle D G F$ are $2 x$ and $2 y$. Since the area of $\triangle B F C$ is $\frac{1}{3}$, we have $3 x+y=\frac{1}{6}$. Similarly, $3 y+2 x=$ the area of $\triangle D E C=\frac{1}{3}$. Solve $x=\frac{1}{42}$ and $y=\frac{2}{21}$. Thus
$$
\frac{\text { Area of } A B G D}{\text { Area of } A B C D}=1-3(x+y)=1-\frac{15}{42}=\frac{9}{14} \text {. }
$$

So $m=9$ and $n=14$. The result follows.