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 ext,∠B and ∠Cntersect at D

We need to prove: `∠D = 1/2 ∠A`

Now, using the exterior angle theorem,

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

\[As \angle \text {ABE  and } \angle \text { ACB  are bisected }\]

\[\angle DCB = \frac{1}{2}\angle ACB\]

Also,

\[\angle DBA = \frac{1}{2}\angle ABE\]

Further, applying angle sum property of the triangle

In  ΔDCB

\[\angle CDB + \angle DCB + \angle CBD = 180^\circ\]

\[ \Rightarrow \angle CDB + \frac{1}{2}\angle ACB + \left( \angle DBA + \angle ABC \right) = 180^\circ\]

\[\angle CDB + \frac{1}{2}\angle ACB + \left( \frac{1}{2}\angle ABE + \angle ABC \right) = 180^\circ . . . . . \left( 2 \right)\]

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

\[\Rightarrow \angle ABC + \angle ABE = 180^\circ\]

\[ \Rightarrow \angle ABC + \frac{1}{2}\angle ABE + \frac{1}{2}\angle ABE = 180^\circ \]

\[ \Rightarrow \angle ABC + \frac{1}{2}\angle ABE = 180^\circ  - \frac{1}{2}\angle ABE . . . . . \left( 3 \right)\]

So, using (3) in (2)

\[\angle CDB + \frac{1}{2}\angle ACB + \left( 180^\circ - \frac{1}{2}\angle ABE \right) = 180^\circ \]

\[ \Rightarrow \angle CDB + \frac{1}{2}\angle ACB - \frac{1}{2}\angle ABE = 0\]

\[ \Rightarrow \angle CDB = \frac{1}{2}\left( \angle ABE - \angle ACB \right)\]

\[ \Rightarrow \angle CDB = \frac{1}{2}\angle CAB\]

\[ \Rightarrow \angle D = \frac{1}{2}\angle A\]

Hence proved.