2009 AMC 10A Problem 23

Add $[ABE]$ to both of the equal areas $[AED]$ and $[BEC]$:
$$[ABD] = [AED] + [ABE] = [BEC] + [ABE] = [ABC].$$

Triangles $ABD$ and $ABC$ share the base $AB$, so $D$ and $C$ are at the same distance from line $AB$. The quadrilateral is convex, so $C$ and $D$ are on the same side of $AB$, which means $CD \parallel AB$: $ABCD$ is a trapezoid.

With $AB \parallel CD$, alternate interior angles give $\angle EAB = \angle ECD$ and $\angle EBA = \angle EDC$, so $\triangle AEB \sim \triangle CED$. The ratio of similarity is
$$\frac{AE}{EC} = \frac{AB}{CD} = \frac{9}{12} = \frac{3}{4}.$$

Since $AE + EC = 14$, we get $AE = \frac{3}{3+4} \cdot 14 = 6$.

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

"Equal areas" for the two triangles on opposite sides of the intersection is the signature of a trapezoid: it is equivalent to $AB \parallel CD$. Once the parallel sides are identified, the diagonals of a trapezoid cut each other in the ratio of the parallel sides. Remember the lemma both ways: equal areas $\Leftrightarrow$ parallel bases.

Let $\theta$ be the angle between the diagonals. Then $[AED] = \frac12 AE \cdot ED \sin\theta$ and $[BEC] = \frac12 BE \cdot EC \sin\theta$, so $AE \cdot ED = BE \cdot EC$, i.e. $\frac{AE}{EC} = \frac{BE}{ED}$. With the vertical angles at $E$ equal, this is SAS similarity of $\triangle AEB$ and $\triangle CED$, giving $\frac{AE}{EC} = \frac{AB}{CD} = \frac34$ directly.

• Trying to solve for AE from the area equation AEED = BEEC alone, instead of deducing the trapezoid structure and using the 3 : 4 similarity.
• Inverting the ratio and taking $AE : EC = 4 : 3$, which gives $AE = 8$ (not a choice) or, after a second slip, one of the wrong fractions.