2018 AMC 10A Problem 24

Quadrilateral $FDBG$ is what is left of $\triangle ABG$ after removing $\triangle ADF$, so compute those two areas.

Triangle $ABG$. By the angle bisector theorem, $\dfrac{BG}{GC} = \dfrac{AB}{AC} = \dfrac{50}{10} = 5$, so $BG = \tfrac56 BC$. Triangles $ABG$ and $ABC$ share the altitude from $A$, so
$$[ABG] = \frac56 [ABC] = \frac56 \cdot 120 = 100.$$

Triangle $ADF$. Since $D$ and $E$ are midpoints, $\triangle ADE$ is $\triangle ABC$ scaled by $\tfrac12$ from $A$, with area $\tfrac14 \cdot 120 = 30$. The same bisector from $A$ cuts $DE$ at $F$, and in $\triangle ADE$ the bisector theorem gives $\dfrac{DF}{FE} = \dfrac{AD}{AE} = \dfrac{25}{5} = 5$, so
$$[ADF] = \frac56 [ADE] = \frac56 \cdot 30 = 25.$$

Therefore $[FDBG] = [ABG] - [ADF] = 100 - 25 = 75$.

The answer is $\boxed{\textbf{(D)}\ 75}$.

Every piece here is an area ratio. The angle bisector theorem converts the side ratio $5:1$ into a base ratio on $BC$, and "same height" converts that into an area ratio. The midsegment $DE$ creates a half-scale copy of the whole configuration (the bisector from $A$ is the same line in both), so the smaller triangle's areas are just one quarter of the larger ones. No lengths, angles or coordinates are ever needed; the given side lengths only matter through their ratio.

Trapezoid $DBCE$ has area $120 - 30 = 90$, and the bisector cuts it into $FDBG$ and $FECG$. Their areas are $[ABG] - [ADF] = \tfrac56(120 - 30) = 75$ and $[ACG] - [AEF] = \tfrac16(120 - 30) = 15$; the same $5:1$ ratio governs both layers, so the quadrilateral is $\tfrac56$ of the trapezoid.

• Forgetting that triangle ADE has one quarter of the area (midpoints halve both sides) or applying the bisector ratio 5:1 to the areas the wrong way round.
• Trying to find the actual angle $A$ from $\sin A = \tfrac{240}{500}$ and computing coordinates, which is far slower and invites arithmetic errors.