15. In the square $A B C D$ below, the area is $1, M$ is the midpoint of side $A D$. Find the area of the shaded part in the figure.
15.【Solution】When utilizing $A M / / B C$, the sides of $\triangle G A M$ and $\triangle G C B$ are proportional.
That is,
thus $\frac{S_{\triangle G G M}}{S_{\triangle G A M}}=\frac{G C}{G A}=2, \frac{S_{\triangle G A B}}{S_{\triangle G A M}}=\frac{G B}{G M}=2$.

Since the side length of the square $A B C D$ is 1, we have
$$
\begin{array}{l}
S_{\triangle A C M}=\frac{1}{2} \times 1 \times \frac{1}{2}=\frac{1}{4}, \\
S_{\triangle A B M}=\frac{1}{2} \times 1 \times \frac{1}{2}=\frac{1}{4},
\end{array}
$$

Therefore, $S_{\triangle G Q M}=\frac{2}{3} S_{\triangle A C M}=\frac{2}{3} \times \frac{1}{4}=\frac{1}{6}$,
$$
\begin{array}{l}
S_{\triangle G A B}=\frac{2}{3} S_{\triangle A B M}=\frac{2}{3} \times \frac{1}{4}=\frac{1}{6} \\
S_{\text {path }}=S_{\triangle G O M}+S_{\triangle G A B}=\frac{1}{6}+\frac{1}{6}=\frac{1}{3}
\end{array}
$$

Thus, the area of the shaded part is $\frac{1}{3}$.