Two Bisectors

#Geometry#AngleBisectors

Problem Statement

In a triangle $ABC, L$ and $K$ are the points of intersections of angle bisectors of $\angle ABC $ and $ \angle BAC $ with segment $AC$ and $BC,$ respectively. The segment $KL$ is the angle bisector of $\angle BLC.$ Determine $ \angle ABC. $

In the diagram above:

Since $AK$ is the angle bisector of $\angle BAC,$ and $LK$ is the angle bisector of $ \angle BLC. $ Therefore, $K$ is the ex-centre of triangle of $ABL.$ Therefore, $BC$ is the angle bisector of the angle $\angle LBD.$

$$ \implies \angle LBC = \angle CBD $$

But we already have

$$ \angle ABL = \angle LBC $$

And further we have

$$ \angle ABL + \angle LBC + \angle CBD = 180 ^\circ \implies \angle ABL = \angle LBC = \angle CBD = 60^\circ $$

$$ \angle ABC = 2 \angle ABL = 120 ^\circ $$