Question

the angle bisectors of $\triangle xyz$ are $\overline{xg}$, $\overline{yg}$, and $\overline{zg}$. they meet at a single point $g$. (in other words, $g$ is the incenter of $\triangle xyz$.) suppose $dg = 12$, $zg = 17$, $m\angle dye = 42^\circ$, and $m\angle fxg = 56^\circ$. find the following measures. note that the figure is not drawn to scale.

Explanation:

Step1: Find $m\angle FXD$

The incenter is equidistant from all sides, so $GD=GF=GE=12$. $\angle GDX = 90^\circ$, $m\angle FXG=56^\circ$. $\angle FXD$ is the sum of $\angle FXG$ and $\angle GXD$. Since $XG$ bisects $\angle FXD$, $\angle FXG=\angle GXD=56^\circ$.
$m\angle FXD = 56^\circ + 56^\circ = 112^\circ$

Step2: Find length of $EG$

The incenter is equidistant from all sides of the triangle, so $EG=DG$.
$EG = DG = 12$

Step3: Find $m\angle FZG$

First, find $\angle XYZ$: $m\angle DYE=42^\circ$, so $m\angle XYZ=42^\circ$. We know $m\angle FXD=112^\circ$, so $m\angle YXZ=112^\circ$. The sum of angles in a triangle is $180^\circ$, so $m\angle YZX = 180^\circ - 112^\circ - 42^\circ = 26^\circ$. $ZG$ bisects $\angle YZX$, so $\angle FZG=\frac{1}{2}m\angle YZX$.
$m\angle FZG = \frac{1}{2} \times 26^\circ = 13^\circ$

Answer:

$m\angle FXD = 112^\circ$
$EG = 12$
$m\angle FZG = 13^\circ$