Angle Bisector Theorem

Question

Let $\triangle ABC$ be a triangle with angle bisector $CL$ $(L\in AB)$ and median $CM$ $(M\in AB)$. I should find $LM$ if $BC=a,AC=b,AB=c(a>b)$. By the angle bisector theorem, $\dfrac{AL}{BL}=\dfrac{AC}{BC}$ or $AL=\dfrac{AC\cdot BL}{BC},BL=\dfrac{AL\cdot BC}{AC}$. We know $LM=AM-AL=BL-BM$ and I am not sure how to approach the problem further.

Answer 1

Since $$\frac{AL}{LB}=\frac{b}{a},$$ we obtain: $$AL=\frac{bc}{a+b}$$ and $$LM=AM-AL=\frac{c}{2}-\frac{bc}{a+b}=\frac{c(a-b)}{2(a+b)}.$$

We can get $AL$ by the following way: $$\frac{AL}{c-AL}=\frac{b}{a}$$ or $$\frac{c-AL}{AL}=\frac{a}{b}$$ or $$\frac{c}{AL}-1=\frac{a}{b}$$ or $$\frac{c}{AL}=1+\frac{a}{b}$$ or $$\frac{c}{AL}=\frac{a+b}{b}$$ or $$\frac{AL}{c}=\frac{b}{a+b}$$ or $$AL=\frac{bc}{a+b}.$$