40 - 120

In triangle $ABC$, the angle $A \hat{B C}$ measures $20^{\circ}$ and the angle $A \hat{C} B$ measures $40^{\circ}$. Let $E$ be a point on $BC$ such that $BE = BA$.

(a) Show that triangle CEA is isosceles.

(b) Knowing that the length of the angle bisector of angle BÂC is 2, determine $\mathrm{BC}-\mathrm{AB}$.
(a) We have $C \hat{A B}=180^{\circ}-20^{\circ}-40^{\circ}=120^{\circ}$. Since triangle $A B E$ is isosceles, it follows that

$$
A \hat{E} B=E \hat{A} B=\frac{180^{\circ}-20^{\circ}}{2}=80^{\circ}
$$

Thus, $\mathrm{CA} E=120^{\circ}-80^{\circ}=40^{\circ}$ and triangle ACE has two angles of $40^{\circ}$, and therefore, is isosceles with $C E=E A$.

(b) Let $\mathrm{D}$ be the foot of the angle bisector of angle BÂC. The bisector divides angle CÂB into two angles of $60^{\circ}$. Therefore, the angle

$$
\mathrm{CDA}=180^{\circ}-40^{\circ}-60^{\circ}=80^{\circ}
$$

Since AÊB also measures $80^{\circ}$, triangle $A D E$ is isosceles. Finally,

![](https://cdn.mathpix.com/cropped/2024_05_01_fcff7d0cdc4488e80dd9g-1.jpg?height=268&width=486&top_left_y=1642&top_left_x=1433)

$$
\mathrm{BC}-\mathrm{AB}=\mathrm{BC}-\mathrm{BE}=\mathrm{CE}=\mathrm{EA}=\mathrm{AD}=2
$$

## Related Problem

The triangle $A B C$ is isosceles with base $B C$ and $B \hat{A C}=48^{\circ}$. Points $D$ and $E$ are on sides $A B$ and $A C$, respectively, such that $D \hat{C A}=9^{\circ}$ and $E \hat{B C}=33^{\circ}$. Determine the measure of angle CDE.

Hint: Use the Pythagorean theorem.

![](https://cdn.mathpix.com/cropped/2024_05_01_fcff7d0cdc4488e80dd9g-2.jpg?height=402&width=474&top_left_y=473&top_left_x=151)

Figure 102.1