Question

ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersect each other at D. Prove that ∠D = $\frac{1}{2}$ ∠A.

Answer 1

In the given ΔABC, the bisectors of and intersect at D

We need to prove:

Now, using the exterior angle theorem,

$\angle ABE=\angle BAC+\angle ACB$ .….(1)

$\mathrm{As}\angle ABE\mathrm{and}\angle ACB\mathrm{are}\mathrm{bisected}$

$\angle DCB=\frac{1}{2}\angle ACB$

Also,

$\angle DBA=\frac{1}{2}\angle ABE$

Further, applying angle sum property of the triangle

In ΔDCB

$$\begin{gathered} \angle CDB+\angle DCB+\angle CBD=180° \\ \Rightarrow \angle CDB+\frac{1}{2}\angle ACB+\left(\angle DBA+\angle ABC\right)=180° \end{gathered}$$

$\angle CDB+\frac{1}{2}\angle ACB+\left(\frac{1}{2}\angle ABE+\angle ABC\right)=180°.....\left(2\right)$

Also, CBE is a straight line, So, using linear pair property

$$\begin{gathered} \Rightarrow \angle ABC+\angle ABE=180° \\ \Rightarrow \angle ABC+\frac{1}{2}\angle ABE+\frac{1}{2}\angle ABE=180° \\ \Rightarrow \angle ABC+\frac{1}{2}\angle ABE=180°-\frac{1}{2}\angle ABE.....\left(3\right) \end{gathered}$$

So, using (3) in (2)

$$\begin{gathered} \angle CDB+\frac{1}{2}\angle ACB+\left(180°-\frac{1}{2}\angle ABE\right)=180° \\ \Rightarrow \angle CDB+\frac{1}{2}\angle ACB-\frac{1}{2}\angle ABE=0 \\ \Rightarrow \angle CDB=\frac{1}{2}\left(\angle ABE-\angle ACB\right) \\ \Rightarrow \angle CDB=\frac{1}{2}\angle CAB \\ \Rightarrow \angle D=\frac{1}{2}\angle A \end{gathered}$$

Hence proved.