2009 AMC 12A 第 20 题

Equal areas of $\triangle AED$ and $\triangle BEC$ imply $\triangle ACD$ and $\triangle BCD$ have equal areas

That forces $AB \parallel CD,$ making $\triangle ABE \sim \triangle CDE$ with ratio $\dfrac{9}{12}$

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.