8. (10 points) In $\triangle A B C$, $B D=D E=E C$, $C F: A C=1: 3$. If the area of $\triangle A D H$ is 24 square centimeters more than the area of $\triangle H E F$, find the area of triangle $A B C$ in square centimeters?
【Answer】Solution: If the area of $\triangle A D H$ is 24 square centimeters more than the area of $\triangle H E F$, then the area of triangle $A D E$ is 24 square centimeters more than the area of triangle $F D E$. Since the area of triangle $F D E$ is equal to the area of triangle $F E C$, this means the area of triangle $A E C$ is 24 square centimeters more than the area of triangle $F E C$. Given that the additional 24 square centimeters is $\frac{2}{3}$ of the area of triangle $A E C$, the area of triangle $A E C$ is $24 \div \frac{2}{3}=36$ square centimeters. Therefore, the area of triangle $A B C$ is $36 \div \frac{1}{3}=108$ (square centimeters). Answer: The area of triangle $A B C$ is 108 square centimeters.