Question

Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12.$ Diagonals $AC$ and $BD$ intersect at $E,$ $AC = 14,$ and $\triangle AED$ and $\triangle BEC$ have equal areas. What is $AE?$

$\dfrac{9}{2}$

$\dfrac{50}{11}$

$\dfrac{21}{4}$

$\dfrac{17}{3}$

$6$

Solution:

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.

2009 AMC 12A Problem 20

Solution:

Problem 20 in Other Years