proof of Stewart’s theorem

Let $\theta$ be the angle $\mathrm{\angle }AXB$.

Cosines law on $\mathrm{△}AXB$ 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 $\mathrm{△}AXC$ and noting that $\psi =\mathrm{\angle }AXC={180}^{\circ }-\theta$ 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

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

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