Prove that $\tan\alpha =\tan^{2}\frac{A}{2}.\tan\frac{B-C}{2}$ Given a triangle ABC with the sides $AB < AC$ and $AM, AD$ respectively median and bisector of angle $A$. Let $\angle MAD = \alpha$. Prove that $$\tan\alpha =\tan^{2}\frac{A}{2}\cdot \tan\frac{B-C}{2}$$

Firstly, in figure 1, 

Using Napier's Analogy in $\Delta ABC$, we have,  
$$\tan \left(\frac{B-C}{2}\right) = \frac{b-c}{b+c} \cdot \cot \left( \frac{A}{2}\right)$$   
Now, in figure 2, 

Using Sine Law in $\Delta ABM$, we have,   
$$\dfrac{x}{\sin \left(\frac{A}{2} + \alpha \right)}=\dfrac{c}{ \sin(\angle BMA)}$$   
Similarly, in  $\Delta ACM$,  
$$\dfrac{x}{\sin \left(\frac{A}{2} - \alpha \right)}=\dfrac{b}{ \sin(\angle CMA)}$$   
Also,  $\sin(\angle CMA)= \sin(\pi-\angle BMA)= \sin(\angle BMA) = \omega $  (let)    
$\therefore b = \dfrac{\omega x}{\sin\left(\frac{A}{2}-\alpha\right)}$    
$\text{and}$  
$c=\dfrac{\omega x}{\sin\left(\frac{A}{2}+\alpha\right)}$   
Putting the values of $b$ and $c$ in the equation of Napier's Analogy, we get (after trivial cancellations),     
$$\tan \left(\dfrac{B-C}{2}\right) = \dfrac{\sin\left(\frac{A}{2}+\alpha\right)-\sin\left(\frac{A}{2}-\alpha\right)}{\sin\left(\frac{A}{2}+\alpha\right)+\sin\left(\frac{A}{2}-\alpha\right)} \cdot \cot \left( \dfrac{A}{2}\right)$$  
$$\implies \tan \left(\dfrac{B-C}{2}\right) = \dfrac{2 \cos\left(\frac{A}{2}\right) \cdot \sin \alpha}{2\sin\left(\frac{A}{2}\right) \cdot \cos \alpha} \cdot \cot \left( \dfrac{A}{2}\right)$$  
$$\implies \tan \alpha = \tan^2 \left( \dfrac{A}{2}\right) \cdot  \tan \left(\dfrac{B-C}{2}\right) $$   
Q.E.D. 
