2009 AMC 10A Problem 23

Problem 23 of 25HarderGeometry

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?

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Solution

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.

More practice

Concepts: triangle area · parallel lines · similarity

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.