In triangle ABC, point D divides side AC so that AD : DC = 1 : 2. Let E be the midpoint of BD and let F be the point of intersection of line BC and line AE. Given that the area of ΔABC is 400 cm2, what is the area of ΔEBA?
If G is the midpoint of DC, the the three triangles BCG, BGD, and BDA have the same area, having the same height BH and equal bases. The two triangles ADE and AEB also have the same area since they have the same height AH' and equal bases. So it follows that the area of AEB is \(\frac{1}{6}\) the area of the triangle ABC, or \(\frac{200}{3}\) square centimeters.
+605 Here's a more braindead but easy solution with mass points. Assign $A=2$ and $C=1$. Then $D=3$, and $B=3$ so $E=6$ then $F=4$. Hence $[AFC]=\frac{800}{3}$, and then $[BEF]=400\cdot 1/3\cdot 1/2=\frac{200}{3}$. So $[EBA]=400-\frac{800}{3}-\frac{200}{3}=\boxed{\frac{200}{3}}$.
