Stewart's Theorem In triangle $ABC$, angle $\angle A = 90^\circ$. Let $M$ be a point on the hypotenuse $BC$. Prove that $MB^2\left(AC^2\right)+MC^2\left(AB^2\right)=MA^2\left(BC^2\right)$. 
Can someone please provide guidance as to how to go about this proof?


Note that $AQ = AC\cdot \frac{BM}{BC}$ and $AP = AB\cdot \frac{CM}{BC}$
Then $AM^2 = AP^2 + AQ^2 = \left(AC\cdot \frac{BM}{BC}\right)^2 + \left(AB\cdot \frac{CM}{BC}\right)^2$  
So $AM^2\cdot BC^2 = BM^2\cdot AC^2 + CM^2\cdot AB^2$ as required.
