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 $AE?$

$\dfrac{9}{2}$

$\dfrac{50}{11}$

$\dfrac{21}{4}$

$\dfrac{17}{3}$

$6$

Answer: E

Concepts:triangle area parallel lines similarity

Difficulty rating: 1690

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.

2009 AMC 10A Problem 23

Answer: E

Difficulty rating: 1690

Solution:

Problem 23 in Other Years