2018 AMC 10A Problem 24

Let $BC = a, BG = x, GC = y,$ and $h$ be the length of the altitude through $A.$ By the Angle Bisector Theorem, $BG:GC=AB:AC=5:1,$ so $BG=\dfrac{5a}{6}.$ Because $DE$ is a midsegment, $DE=\dfrac a2.$ Applying the same theorem in $\triangle ADE$ gives $DF:FE=AD:AE=5:1,$ so $DF=\dfrac56DE=\dfrac{5a}{12}.$ The trapezoid’s height is $\frac{h}{2},$ and $\frac{ah}{2}=120.$ Its average base length is $\dfrac12\left(\dfrac{5a}{12}+\dfrac{5a}{6}\right)=\dfrac{5a}{8}.$ Therefore, its area is $$\dfrac{5a}{8}\cdot\dfrac h2=\dfrac58\cdot\dfrac{ah}{2}=75.$$ Thus, D is the correct answer.