Prove that $CR$ bisects $\angle C$

Question

The external bisector of $\angle A$ of triangle $ABC$ meets $BC$ produced at L and the internal bisector of $\angle B$ meets $CA$ at $M$ if $LM$ meets $AB$ at $R$ prove that $CR$ bisects angle $C$. This problem is from an excursion in mathematics, section: concurrency and colinearity

My Attempt: Since in this section the Ceva's and Menelaus theorem were given I am thinking we somehow need to use them to get $AR$/$BR$ = $AC$/$BC$ so I applied it to various triangle but did not get anything new . I also got angle C = 2a -2b where a and b are the angles bisected of exterior angle A and interior angle B respectively

Answer 1

A very simple approach based on the law of sines is as follows:

In $\triangle LBA$, we have:

$$\frac{AR}{BR}=\frac{\sin \angle ALR}{\sin \angle BLR} \times \frac{\sin (180^{\circ}-\angle B)}{\sin (90^{\circ}-\frac{\angle A}{2})}=\frac{\sin \angle ALR}{\sin \angle BLR} \times\frac{\sin \angle B}{\cos \frac{\angle A}{2}}.$$

On the other hand, in $\triangle LMA$, we have:

$$\frac{\sin \angle ALR}{\sin (90^{\circ}+\frac{\angle A}{2})}=\frac{AM}{LM},$$ and, in $\triangle LCM$:

$$\frac{\sin \angle BLR}{\sin \angle C)}=\frac{MC}{LM}.$$

Therefore, we will get that:

$$\frac{AR}{BR}=\frac{AM}{MC}\times \frac{\sin (90^{\circ}+\frac{\angle A}{2})}{\sin \angle C}\times \frac{\sin \angle B}{\cos \frac{\angle A}{2}}=\frac{AM}{MC}\times \frac{AC}{AB}=\frac{AB}{BC}\times \frac{AC}{AB}=\frac{AC}{BC},$$

which proves the claim of the problem. Note that $\frac{AM}{MC}=\frac{AB}{BC}$ is concluded from the fact that $BM$ is the angle bisector.

We are done.

Answer 2

Here's a diagram I constructed on GeoGebra based on the question: diagram

We're trying to prove the red line ($CR$) bisects angle $C$. Say $CR$ intersects line $AL$ at $X$. By the converse of the internal angle bisector theorem, if we prove that $\frac{AX}{XL} = \frac{AC}{LC}$, then we have proven that $CR$ bisects $\angle{C}$. From the question and represented diagram, we know that the cevians of $\triangle{ALC}$ are collinear at point $R$, and we can apply Ceva's theorem on $\triangle{ALC}$.

Applying Ceva's theorem on $\triangle{ALC}$,

$\frac{AX}{XL} \cdot \frac{LB}{BC} \cdot \frac{CM}{MA} = 1$

Also, from the internal angle bisection of $BM$ of $\angle{B}$, by using the angle bisector theorem again we get $\color{green}{\frac{CM}{MA} = \frac{BC}{AB}} $ (you can see this by tilting your head to an angle seeing $AC$ as the base of $\triangle{ABC}$)

Now, from the external angle bisector theorem, with $AL$ bisecting $\angle{A}$, $\frac{LB}{LC} = \frac{AB}{AC}$. Rearranging, we get $\color{blue}{\frac{LB}{AB} = \frac{LC}{AC}}$. Note: I'm considering all the segments' taken to be absolute values.

Substituting the green statement into the statement from Ceva's theorem, we get

$\frac{AX}{XL} \cdot \frac{LB}{BC} \cdot \color{green}{\frac{BC}{AB}} = 1$. Simplifying, $\frac{AX}{XL} \cdot \frac{LB}{AB} = 1$. Substituting the blue statement,$\frac{AX}{XL} \cdot \color{blue}{\frac{LC}{AC}} = 1$.

Therefore, we get $\frac{AX}{XL} = \frac{AC}{LC}$, proving what we originally wanted to prove so by the converse of the angle bisector theorem, we have proven $CR$ bisects $\angle{C}$.