2018 AMC 10A Problem 24

The quadrilateral $FDBG$ is triangle $ABG$ with triangle $ADF$ cut off the top. So find those two areas.

1. How the bisector splits area. The point $G$ is on the bisector of angle $A$, so it is the same distance from line $AB$ as from line $AC$. Call that distance $h$. Then
$$[ABG] = \tfrac12 \cdot AB \cdot h, \qquad [ACG] = \tfrac12 \cdot AC \cdot h.$$
Their ratio is $AB : AC = 50 : 10 = 5 : 1$.
2. Area of $ABG$. Triangle $ABC$ (area $120$) is split into $6$ equal shares, $5$ for $ABG$ and $1$ for $ACG$:
$$[ABG] = \tfrac56 \cdot 120 = 100.$$
3. Area of $ADE$. $D$ and $E$ are midpoints, so $AD = 25$ and $AE = 5$. Triangle $ADE$ is triangle $ABC$ shrunk to half size, and halving every length divides the area by $2 \cdot 2 = 4$. So $[ADE] = \tfrac{120}{4} = 30$.
4. Area of $ADF$. The same bisector passes through $F$ on $DE$. By the same reasoning as in step 1, it splits triangle $ADE$ in the ratio $AD : AE = 25 : 5 = 5 : 1$. So
$$[ADF] = \tfrac56 \cdot 30 = 25.$$
5. Subtract. $[FDBG] = [ABG] - [ADF] = 100 - 25 = 75$.

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

Any point on an angle bisector is equally far from the two sides of the angle, so the two triangles it creates share a height and their areas follow the ratio of the sides. The midpoints make a half-size copy of the whole picture with the same bisector, so the same $5:1$ split repeats inside it. The actual lengths and angles never need to be found.

Look only at the bottom part. The trapezoid $DBCE$ has area $120 - 30 = 90$. The bisector splits the whole triangle $5 : 1$ and the top triangle $5 : 1$, so it also splits what is left, the trapezoid, $5 : 1$. The part on the $B$ side is $FDBG$, with area $\tfrac56 \cdot 90 = 75$.

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 $\boxed{\textbf{(D)}\ 75}$.