In the following drawing, $\angle C B G=20^{\circ}, \angle G B E=40^{\circ}, \angle E B F=20^{\circ}, \angle B C F=50^{\circ}$ and $\angle F C E=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-11.jpg?height=603&width=703&top_left_y=1776&top_left_x=759)
a) Verify that $B G=B F$.

b) Verify that $F G=E G$.

c) Find the value of the measure of angle $x$.
Solution

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-12.jpg?height=613&width=727&top_left_y=738&top_left_x=543)

a) We have

$$
\angle C G B=180^{\circ}-\angle B C G-\angle C B G=80^{\circ}=\angle B C G
$$

Therefore, $B C=B G$. On the other hand,

$$
\angle B F C=180^{\circ}-\angle C B F-\angle B C F=50^{\circ}=\angle B C F
$$

thus $B C=B F$. Hence,

$$
B G=B C=B F
$$

b) Since $B G=B F$ and $\angle G B F=60^{\circ}$, triangle $B F G$ is isosceles with base angles given by

$$
\frac{180^{\circ}-\angle G B F}{2}=60^{\circ}
$$

That is, $B F G$ is an equilateral triangle, and thus $F G=B G$. Since

$$
\angle B G E=180^{\circ}-\angle B G C=100^{\circ}
$$

it follows that

$$
\angle G E B=180^{\circ}-\angle B G E-\angle G B E=40^{\circ}
$$

and thus $\triangle B E G$ is isosceles with side $E G=B G=F G$.
c) Since $\angle F G E=180^{\circ}-\angle B G F-\angle B G C=40^{\circ}$ and $F G=B G=E G$, it follows that triangle $E F G$ is isosceles with base angle given by

$$
\frac{180^{\circ}-40^{\circ}}{2}=70^{\circ}
$$

Thus, $x+40^{\circ}=70^{\circ}$, or $x=30^{\circ}$.