2009 AMC 10A Problem 23

Attempt Problem 23 of the 2009 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

23.

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

Answer: E

Concepts:triangle area parallel lines similarity

Difficulty rating: 1690

Small hint:

Adding

\triangle CED

to each equal-area triangle shows

\triangle ACD

and

\triangle BCD

have equal areas

Big hint:

Equal areas over the shared base

CD

force

AB \parallel CD,

making

\triangle ABE \sim \triangle CDE

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.