We have a $\triangle ABC$ with $\angle ACB=120^\circ$. $CL$ and $AT$ are angle bisectors, $L\in AB$ and $T\in BC$. I should show that $LT$ is the angle bisector of $\angle CLB$.
By the angle bisector theorem, $\dfrac{AL}{LB}=\dfrac{AC}{BC}$ and $\dfrac{CT}{TB}=\dfrac{AC}{AB}$. We should show $\dfrac{CT}{TB}=\dfrac{CL}{LB}$. How can I use $\angle ACL=\angle BCL=60^\circ$? Thank you in advance!
Best Answer
Observe a triangle $ACL$ and extend halfline $AC$ to $X$.
Notice that $CT$ is outer angle bisector for angle $\angle ACL$ since $\angle LCT = \angle LCX = 60^{\circ}$.
Since $T$ is also on angle bisector $AT$ of angle $\angle CAL$. So $T$ is at equal distance to lines $CL$ and $LB$, so $LT$ is angle bisector of angle $\angle CLB$.