proof of Stewart’s theorem

Let $\theta$ be the angle $\angle AXB$ .

Cosines law on $\triangle AXB$ says $c^{2}=m^{2}+p^{2}-2pm\cos\theta$ and thus

$\cos\theta=\frac{m^{2}+p^{2}-c^{2}}{2pm}$

Using cosines law on $\triangle AXC$ and noting that $\psi=\angle AXC=180^{\circ}-\theta$ and thus $\cos\theta=-\cos\psi$ we get

$\cos\theta=\frac{b^{2}-n^{2}-p^{2}}{2pn}.$

From the expressions above we obtain

$2pn(m^{2}+p^{2}-c^{2})=2pm(b^{2}-n^{2}-p^{2}).$

By cancelling $2p$ on both sides and collecting we are led to

$m^{2}n+mn^{2}+p^{2}n+p^{2}m=b^{2}m+c^{2}n$

and from there $mn(m+n)+p^{2}(m+n)=b^{2}m+c^{2}n$ . Finally, we note that $a=m+n$ so we conclude that

$a(mn+p^{2})=b^{2}m+c^{2}n.$

QED