AMC10 2009 10A, Problem 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 $AE$?

Solution: Let $[ABC]$ denote the area of triangle $ABC$. $[AED] = [BEC]$, so $[ABD] = [AED] + [AEB] = [BEC] + [AEB] = [ABC]$. Since triangles $ABD$ and $ABC$ share a base, they also have the same height and thus $\overline{AB}||\overline{CD}$ and $\triangle{AEB}\sim\triangle{CED}$ with a ratio of $3: 4$. $AE = \frac {3}{7}\times AC$, so $AE = \frac {3}{7}\times 14 = 6\ \fbox{6}$. [asy]pathpen = linewidth(0.7);pointpen = black; pair D=MP("D",(0,0)),C=MP("C",(12,0)),A=MP("A",C+14*expi(145*pi/180),N),B=MP("B",A+(9,0),N),E=IP(A--C,B--D);MP("9",(A+B)/2,N);MP("12",(C+D)/2); fill(A--D--E--cycle,rgb(0.8,0.8,0.8));fill(B--C--E--cycle,rgb(0.8,0.8,0.8));D(A--B--C--D--cycle);D(A--C);D(B--D);D(E); [/asy]

Answer: 6