Question

prove that the angle between internal bisector of one base angle and external bisector of other base angle of a triangle equal to one half of the vertical angle

Answer 1

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

## Solution:

Draw a triangle ABC.

Let B and C are base angles and A be the vertical angle.

Draw BX internal bisector of angle B.

Draw CY external bisector of angle C.

Let BX and CY intersect at D.

Now $<\mathrm{BDC}$ is the angle between internal bisector of one base angle (B) and external bisector of other base angle (C) of triangle ABC.

From triangle DBC.

$$< B D C = 1 8 0 - (< D B C + < D C B)$$

Where $<\mathrm{DBC}$ is internally bisected angle of B and $<\mathrm{DCB}$ is $<\mathrm{C} +$ externally bisected angle of C

$$< B D C = 1 8 0 - \left\lfloor < \left(\frac {B}{2}\right) + < C + \text{exterior angle of } \frac {C}{2} \right\rfloor$$

Recall that exterior angle of C equal $(< A + < B)$

So

$$\begin{array}{l} < B D C = 1 8 0 - \left\lfloor < \left(\frac {B}{2}\right) + < C + < \left(\frac {A + B}{2}\right) \right\rfloor \\ < B D C = 1 8 0 - \left\lfloor < B + < C + < \left(\frac {A}{2}\right) \right\rfloor \\ < B D C = \lfloor 1 8 0 - (< B + < C) \rfloor - < \left(\frac {A}{2}\right) \\ < B D C = < A - < \left(\frac {A}{2}\right) \\ < B D C = < \left(\frac {A}{2}\right) \\ \end{array}$$

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