1. As shown in Figure $2, A D$
$/ / B C$, trapezoid $A B C D$
has an area of $180, E$ is
the midpoint of $A B, F$ is a point
on side $B C$, and $A F$
$/ / D C, A F$ intersects $E D$ and $B D$ at points $G$ and $H$. Let $\frac{B C}{A D}=$ $m(m \in \mathbf{N})$. If the area of $\triangle G H D$ is an integer, then the value of $m$ is $\qquad$
2, 1.2 or 5.
As shown in Figure 6, draw $B K / / A F$ intersecting $E D$ at point $K$, then $\triangle K E B \cong \triangle G E A$
$$
\begin{array}{l}
\text { Hence } \frac{G H}{A G}=\frac{G H}{B K} \\
=\frac{H D}{B D}=\frac{F C}{B C} \\
=\frac{A D}{B C}=\frac{1}{m} .
\end{array}
$$

Thus, we have $S_{\triangle A B D}=\frac{1}{m+1} S_{\text {quadrilateral } A B C D}=\frac{180}{m+1}$,
$$
\begin{array}{l}
S_{\triangle A R D}=\frac{1}{m} S_{\triangle A R D}=\frac{180}{m(m+1)}, \\
S_{\triangle C A D}=\frac{1}{m+1} S_{\triangle A H D}=\frac{180}{m(m+1)^{2}} .
\end{array}
$$

It is easy to see that $\frac{180}{m(m+1)^{2}}$ is an integer, so, $(m+1)^{2} \backslash 180$.
Since $180=2^{2} \times 3^{2} \times 5$, we have $m+1=2,3$ or 6. After verification, $m+1=3$ or 6.