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$ABC$ is a triangle. The bisector of the exterior angle at $B$ and the bisector of $\angle C$ intersect each other at $D$. Prove that $\angle D = \frac{1}{2}\angle A$.
Given:
$ABC$ is a triangle. The bisector of the exterior angle at $B$ and the bisector of $\angle C$ intersect each other at $D$.
To do:
We have to prove that $\angle D = \frac{1}{2}\angle A$.
Solution:
In $\triangle ABC, CB$ is produced to $E$.
The bisectors of $\angle ABE$ and $\angle ACB$ meet at $D$.
In $\triangle \mathrm{BDC}$,
$\angle \mathrm{ABE}=\angle \mathrm{A}+\angle \mathrm{C}$
$\frac{1}{2} \angle \mathrm{ABE}=\frac{1}{2} \angle \mathrm{A}+\frac{1}{2} \angle \mathrm{C}$
$\angle 1=\frac{1}{2} \angle \mathrm{A}+\angle 4$..............(i)
In $\triangle \mathrm{BDC}$,
$\angle 2=\angle \mathrm{D}+\angle 4$
$\angle \mathrm{D}=\angle 2-\angle 4$
$ = \angle 1-\angle 4$
$=(\frac{1}{2} \angle \mathrm{A}+\angle 4)-\angle 4$ [From (i)]
$=\frac{1}{2} \angle \mathrm{A}+\angle 4-\angle 4$
$=\frac{1}{2} \angle \mathrm{A}$
Hence, $\angle \mathrm{D}=\frac{1}{2} \angle \mathrm{A}$.
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