Question

Triangle $ABC$ with $AB = 50$ and $AC = 10$ has area $120.$ Let $D$ be the midpoint of $\overline{AB},$ and let $E$ be the midpoint of $\overline{AC}.$ The angle bisector of $\angle BAC$ intersects $\overline{DE}$ and $\overline{BC}$ at $F$ and $G,$ respectively. What is the area of quadrilateral $FDBG?$

$60$

$65$

$70$

$75$

$80$

Solution:

Since $D$ and $E$ are midpoints, $\triangle ADE$ has $\tfrac14$ the area of $\triangle ABC,$ namely $30,$ so trapezoid $EDBC$ has area $120 - 30 = 90.$

By the Angle Bisector Theorem, $G$ divides $BC$ with $BG = \tfrac{AB}{AB + AC} \cdot BC = \tfrac56 BC,$ and likewise $F$ divides $DE$ so that $DF = \tfrac56 DE.$ Because $FDBG$ and $EDBC$ share the same height, the area of $FDBG$ is $\tfrac56$ of the area of $EDBC:$ $\tfrac56 \cdot 90 = 75.$

Thus, the correct answer is D.

2018 AMC 12A Problem 18

Solution:

Problem 18 in Other Years