## Area in Right Triangle

In the diagram, $\angle{ABC}$ is a right angle. Point $D$ is on $\overline{BC}$, and $\overline{AD}$ bisects $\angle{CAB}$. Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$, respectively, so that $AE=3$ and $AF=10$. Given that $EB=9$ and $FC=27$, find the integer closest to the area of quadrilateral $DCFG$.

Source: NCTM Mathematics Teacher

SOLUTION
Angle Bisector Theorem
The angle bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.

$\frac{DB}{DC}=\frac{AB}{AC}$

Divided Triangle Proposition
When a triangle is divided into two smaller triangles by a segment from one of its vertices to the opposite side (whether or not this segment is an angle bisector), the ratio of the areas of the two smaller triangles is equal to the ratio of the triangles’ bases.

$\frac{area\;ADB}{area\;ACD}=\frac{DB}{DC}$

By the Pythagorean theorem
$AB^2+BC^2=AC^2$
$12^2+BC^2=37^2$
$BC^2=\sqrt{1225}$
$BC=35$

$Area\;DCFG=area\;ADC-area\;AGF$
Calculate the area of $ADC$
$\overline{AD}$ is the angle bisector of $\angle{A}$ of $\triangle{ABC}$. By the Angle Bisector theorem
$\frac{DC}{DB}=\frac{AC}{AB}=\frac{37}{12}$
By the Divided Triangle proposition
$\frac{area\;ADC}{area\;ABD}=\frac{DC}{DB}=\frac{37}{12}$
$\frac{area\;ADC}{37}=\frac{area\;ABD}{12}=\frac{area\;ADC+\;area\;ABD}{37+12}=\frac{area\;ABC}{49}$
$area\;ADC=\frac{37}{49}\left (area\;ABC\right )$
$area\;ABC=\frac{1}{2}\left (BC\times BA\right )$
$=\frac{1}{2}\left (35\times 12\right )$
$=210$
$area\;ADC=\frac{37}{49}\left (210\right )$
$=158.57$
Calculate the area of $AGF$
$\overline{AG}$ is the angle bisector of $\angle{A}$ of $\triangle{FAE}$. By the Angle Bisector theorem
$\frac{GF}{GE}=\frac{AF}{AE}=\frac{10}{3}$
By the Divided Triangle proposition
$area\;AGF=\frac{10}{13}\left (area\;FAE\right )$

$\overline{FE}$ is a segment from vertex $F$ to point $E$ on opposite side $\overline{AB}$ in $\triangle{FAB}$
$area\;FAE=\frac{3}{12}\left (area\;FAB\right )$

$\overline{BF}$ is a segment from vertex $B$ to point $F$ on opposite side $\overline{AC}$ in $\triangle{ABC}$
$area\;FAB=\frac{10}{37}\left (area\;ABC\right )$

Finally
$area\;AGF=\frac{10}{13}\left (area\;FAE\right )$
$=\frac{10}{13}\left (\frac{3}{12}area\;FAB\right )$
$=\frac{10}{13}\cdot\frac{3}{12}\left (\frac{10}{37}area\;ABC\right )$
$=\frac{10}{13}\cdot\frac{3}{12}\cdot\frac{10}{37}\left (210\right )$
$=10.91$
$Area\;DCFG=area\;ADC-area\;AGF$
$=158.57-10.91$
$=147.66$
Integer closest to $147.66$ is $148$.

Answer: $148$ square units