14. (12 points) In a square $A B C D$ with an area of 360, $E$ is the midpoint of $A D$, $H$ is the midpoint of $F G$, and $D F=C G$. The area of triangle $A G H$ is ( )
A. 70
B. 72
C. 75
D. 90
【Solution】Solution: According to the analysis, connect $E G, E F$, let the side length of the square be 1 unit, $G C=D F=x$ units.
By the kite model, we know that the ratio of areas is: $S_{\triangle E G C}: S_{\triangle E C F}=G H: H F=1: 1, x \times 1=(1-x) \times \frac{1}{2}$,
Solving this, we get: $x=\frac{1}{3}$.
Connect $A F, \because S_{\triangle \mathrm{ADF}}=\frac{1}{3} \times 1 \times \frac{1}{2}=\frac{1}{6} ; S_{\triangle G C F}=\frac{1}{2} \times \frac{1}{3} \times\left(1-\frac{1}{3}\right)=\frac{1}{9}$;
$S_{\triangle A B G}=\frac{1}{2} \times 1 \times\left(1-\frac{1}{3}\right)=\frac{1}{3}$,
$\therefore S_{\triangle A G F}=1-S_{\triangle A B G}-S_{\triangle G C F}-S_{\triangle A D F}=1-\frac{1}{3}-\frac{1}{9}-\frac{1}{6}=\frac{7}{18}$,
$\therefore S_{\triangle A G H}=\frac{1}{2} S_{\triangle A G F}=\frac{1}{2} \times \frac{7}{18} \times 360=70$.
Therefore, the answer is: $A$.