2009 AMC 12A Problem 20

Adding $\triangle CED$ to each of $\triangle AED$ and $\triangle BEC$ shows $\triangle ACD$ and $\triangle BCD$ have equal areas. They share base $CD,$ so $A$ and $B$ are equidistant from line $CD,$ meaning $AB \parallel CD.$ Then $\triangle ABE \sim \triangle CDE$ with ratio $\dfrac{AB}{CD} = \dfrac{9}{12} = \dfrac{3}{4},$ so $\dfrac{AE}{EC} = \dfrac{3}{4}.$ Writing $AE = 3x$ and $EC = 4x,$ we get $7x = AC = 14,$ so $x = 2$ and $AE = 6.$ Thus, the correct answer is E.