2009 AMC 10A Problem 23

Since $[AED] = [BEC],$ adding $[CED]$ to both gives $[ACD] = [BCD].$ These share base $CD,$ so $A$ and $B$ are equidistant from line $CD,$ meaning $AB \parallel CD.$ Then $\triangle ABE \sim \triangle CDE$ with ratio $\dfrac{AB}{CD} = \dfrac{9}{12} = \dfrac34,$ so $\dfrac{AE}{EC} = \dfrac34.$ With $AE + EC = AC = 14,$ we get $AE = \dfrac{3}{7} \cdot 14 = 6.$ Thus, the correct answer is E.