14. (6 points) As shown in the figure, in triangle $A B C$, the length of segment $E C$ is twice the length of segment $B E$, and the length of segment $C D$ is twice the length of segment $A D$. Given that the area of triangle $B D E$ is 14 square centimeters, what is the area of triangle $A B C$ in square centimeters?
【Solution】Solution: According to the problem, the length of line segment $BC$ is 3 times the length of line segment $BE$, and the height of triangle $ABC$ is $\frac{3}{2}$ times the height of triangle $BDE$,
$\therefore$ the area of triangle $ABC$ is $\frac{9}{2}$ times the area of triangle $BDE$,
$\because$ the area of triangle $BDE$ is 14 square centimeters,
$\therefore$ the area of triangle $ABC$ is $14 \times \frac{9}{2}=63$ square centimeters,
Answer: The area of triangle $ABC$ is 63 square centimeters.