9. In the right figure, $\mathrm{ABCD}$ is a parallelogram, $\mathrm{E}$ is the midpoint of $\mathrm{CD}$, the intersection point of $\mathrm{AE}$ and $\mathrm{BD}$ is $\mathrm{F}$, the intersection point of $\mathrm{AC}$ and $\mathrm{BE}$ is $\mathrm{H}$, the intersection point of $\mathrm{AC}$ and $\mathrm{BD}$ is $\mathrm{G}$, the area of quadrilateral EHGF is 15 square centimeters, then the area of $\mathrm{ABCD}$ is $\qquad$ square centimeters.
Analysis: Examining several major geometric models.
Solution one: Butterfly model and half model.
(1) $E$ is the midpoint of $C D$, $D E: A B=1: 2$, so $S_{\triangle D E E}: S S_{D A F}: S_{\triangle B E F}: S_{\triangle A B F}=1: 2: 2: 4$.
(2) Let the area of the parallelogram be “1”. Since $E$ is the midpoint of $C D$, $\mathrm{S}_{\triangle \triangle B G} 、 \mathrm{~S}_{\triangle A D G} 、 \mathrm{~S}_{\triangle B E C}$ each occupy $\frac{1}{4}$ of the parallelogram's area, and trapezoid $\mathrm{S}_{\text {ABED }}$ occupies $\frac{3}{4}$ of the parallelogram's area;
(3) Therefore, $S \triangle_{D: R}=\frac{3}{4} \times \frac{2}{1+2+2+4}=\frac{1}{6}, S \triangle_{\mathrm{GAF}}=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$, similarly, $S \triangle_{G \mathrm{GIB}}=\frac{1}{12}$.
(4) According to the half model, $S_{\triangle \mathrm{ABE}}=\frac{1}{2}, S_{\text {quadrilateral }}=\frac{1}{2}-\frac{1}{4}-\frac{1}{12}-\frac{1}{12}=\frac{1}{12}$;
(5) The area of $\mathrm{ABCD}$ is $15 \div \frac{1}{12}=180 \mathrm{~cm}^{2}$.

Solution two: Similarity model, equal area transformation, and half model.
(1) $E$ is the midpoint of $C D$, $D E: A B=1: 2$, so $D F: F B=1: 2$, and $D G=G B$,
$$
\mathrm{DF}: \mathrm{FG}=\frac{1}{1+2}:\left(\frac{1}{2}-\frac{1}{1+2}\right)=2: 1 \text {; }
$$
(2) Let the area of the parallelogram be “1”. Since $E$ is the midpoint of $\mathrm{CD}$, $\mathrm{S}_{\triangle A B B} 、 \mathrm{~S}_{\triangle A D C}$ each occupy $\frac{1}{4}$ of the parallelogram's area, so $\mathrm{S} \triangle_{\mathrm{GAR}}=\frac{1}{4} \times \frac{1}{2+1}=\frac{1}{12}$, similarly, $\mathrm{S} \triangle_{\mathrm{GHBB}}=\frac{1}{12}$.
(3) According to the half model, $\mathrm{S}_{\triangle \mathrm{ABE}}=\frac{1}{2}, \mathrm{~S}_{\text {quadrilateral BIGFF }}=\frac{1}{2}-\frac{1}{4}-\frac{1}{12}-\frac{1}{12}=\frac{1}{12}$;
(4) The area of $\mathrm{ABCD}$ is $15 \div \frac{1}{12}=180 \mathrm{~cm}^{2}$.

Solution three: Swallowtail model and half model.
(1) Let the area of the parallelogram be “1”. $\mathrm{S}_{\triangle A D C}=\frac{1}{2}$.
(2) $E$ is the midpoint of $C D$, $G$ is the midpoint of $A C$, connect $\mathrm{FC}$, let $\mathrm{S}_{\triangle D E F}$ be 1 part, $S_{\triangle}$ BCF is also 1 part, according to the swallowtail, $\mathrm{S}_{\triangle A D F}$ is 2 parts, and according to the swallowtail, $\mathrm{S}_{\triangle A C F}$ is also 2 parts, according to the proportional distribution, $\mathrm{S}_{\triangle \triangle G F} 、 \mathrm{S}_{\triangle G C F}$ are both 1 part, so $\mathrm{S} \triangle_{\mathrm{GAF}}=\frac{1}{2} \div(2+1+1+1+1)$ $=\frac{1}{12}$, similarly, $\mathrm{S} \triangle_{\mathrm{GHB}}=\frac{1}{12}$.
(3) According to the half model, $\mathrm{S}_{\triangle \mathrm{ABE}}=\frac{1}{2}$, $\mathrm{S}_{\text {quadrilateral BigF }}=\frac{1}{2}-\frac{1}{4}-\frac{1}{12}-\frac{1}{12}=\frac{1}{12}$;
(4) The area of $\mathrm{ABCD}$ is $15 \div \frac{1}{12}=180 \mathrm{~cm}^{2}$.

Solution four: Kite model and half model.
Connecting EG can also solve the problem.