70. As shown in the figure, quadrilateral $A B C D$ is a rectangle, $E, F$ are points on $A B, B C$ respectively, and $A E=\frac{1}{3} A B$, $C F=\frac{1}{4} B C, A F$ intersects $C E$ at $G$. If the area of rectangle $A B C D$ is 120, then the sum of the areas of $\triangle A E G$ and $\triangle C G F$ is $\qquad$
answer: 15