Question

$ABC$ is a triangle. The bisectors of the internal angle $\angle B$ and external angle $\angle C$ intersect at $D.$ if $\angle BDC={60}^{\circ }$ then $\angle A$ is

• A. ${120}^{\circ }$
• B. ${180}^{\circ }$
• C. ${60}^{\circ }$
• D. ${150}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

Consider $△ABC$

Let $BC$ be extended to $E$

Since Angular bisectors Meet at $D$

$\angle ABD=\angle DBC\cdots \left(1\right)$

$\angle ACD=\angle DCE\cdots \left(2\right)$

Consider $△DBC$

By External sum property

$\angle DCE=\angle BDC+\angle DBC$

$⟹2\angle DCE=2\left({60}^{\circ }\right)+2\angle DBC$

$⟹\angle ACE={120}^{\circ }+\angle ABC$

By external sum property of $△ABC$

$\angle ACE=\angle BAC+\angle ABC$

$⟹\angle A={60}^{\circ }$