Question

Prove that the angle between the internal bisector of one base angle and the external bisector of the other is equal to one half of the vertical angle.

Answer 1

Applying the exterior angle property:

The measure of the exterior angle of a triangle equals the sum of the two opposite angles of the triangle.

Therefore,

$$\begin{gathered} In△ABC, \\ \angle A+\angle B=\angle ACD \\ Multiplyingby\frac{{\displaystyle 1}}{{\displaystyle 2}}, \\ \Rightarrow \frac{1}{2}\angle A+\frac{1}{2}\angle B=\frac{1}{2}\angle ACD \\ \Rightarrow \frac{1}{2}\angle A+\angle x=\angle y...\left(1\right) \\ In△BCE, \\ \angle x+\angle E=\angle ECD \\ \Rightarrow \angle x+\angle E=\angle y...\left(2\right) \end{gathered}$$

Comparing the equations:

$$\begin{gathered} \angle x+\angle E=\angle x+\frac{1}{2}\angle A \\ \Rightarrow \angle E=\frac{1}{2}\angle A \end{gathered}$$

Hence, it is proved that the angle between the internal bisector of one base angle and the external bisector of the other is equal to one half of the vertical angle.