If $M\in BC$ of $\triangle ABC$, and $2\cot \angle AMC =\cot B-\cot C$, then $AM$ is a median

Question

In $\triangle ABC$, with $M\in BC$ and $\angle AMC=\theta$, if we have $$2\cot \theta =\cot B-\cot C$$ then we may conclude that $AM$ is a median.

My try :

$AH=h$ altitude , then $BM=x$ and $CM=y$ ,then $MH=z$ , $HC=w$

$CM=CH+HM=x$ so we will prove that $x=y$

$$\cot \theta =\frac{z}{h}$$

$$\cot B=\frac{x+z}{h}$$

$$\cot C =\frac{w}{h}$$

Then from first rotation we concluded !

$2\frac{z}{h}=\frac{x+z-w}{h}$

This mean :

$x-w=z\implies x=w+z\implies x=y$

I'm correct or no !?

Answer 1

Here is an alternative, maybe cleaner, proof. Apply the sine rule to the triangles ABM and ACM,

$$\frac{BM}{CM} = \frac{\frac{BM}{AM}}{\frac{CM}{AM}} =\frac{\frac{\sin(\theta-B)}{\sin B}}{\frac{\sin(\theta+C)}{\sin C}} =\frac{\sin\theta\cot B - \cos\theta}{\sin\theta\cot C + \cos\theta} =\frac{\cot B - \cot\theta}{\cot C + \cot\theta}=1 $$