2009 AMC 10A Problem 23

1. The equal areas are $[AED]$ and $[BEC]$ (square brackets mean area). Add the small triangle $ABE$ to each:
- $[AED] + [ABE]$ is the area of triangle $ABD$.
- $[BEC] + [ABE]$ is the area of triangle $ABC$.
So triangles $ABD$ and $ABC$ have the same area.
2. These two triangles have the same base $AB$. Equal area and equal base means equal height, so $D$ and $C$ are the same distance from line $AB$. Both are on the same side of $AB$ (the shape is convex), so $CD$ is parallel to $AB$. The quadrilateral is a trapezoid.
3. Because $AB$ and $CD$ are parallel, triangles $AEB$ and $CED$ have matching angles (a diagonal crossing the two parallel sides makes equal alternate angles, and the angles at $E$ are opposite each other). So they have the same shape, and triangle $CED$ is triangle $AEB$ scaled up by $\frac{12}{9} = \frac{4}{3}$.
4. That means $AE$ and $EC$ are in the ratio $3$ to $4$. Split $AC = 14$ into $3 + 4 = 7$ equal parts of $2$:
$$AE = 3 \times 2 = 6, \qquad EC = 4 \times 2 = 8.$$

The answer is $\boxed{\textbf{(E)}\ 6}$.

In a quadrilateral cut by its diagonals, triangles $AED$ and $BEC$ (the ones touching sides $AD$ and $BC$) have equal areas exactly when $AB$ is parallel to $CD$. Once you know it is a trapezoid, the diagonals cut each other in the ratio of the parallel sides. Remember this both ways: equal areas mean parallel sides, and parallel sides mean equal areas.

Use the area formula with an angle. If the diagonals cross at angle $\theta$, then $[AED] = \frac12 \cdot AE \cdot ED \cdot \sin\theta$ and $[BEC] = \frac12 \cdot BE \cdot EC \cdot \sin\theta$ (the angles at $E$ in these two triangles are equal, being opposite each other). Equal areas give $AE \cdot ED = BE \cdot EC$, so $\frac{AE}{EC} = \frac{BE}{ED}$. Together with the equal angles at $E$, triangles $AEB$ and $CED$ are similar, so $\frac{AE}{EC} = \frac{AB}{CD} = \frac{3}{4}$, and $AE = 6$.