One way to prove this theorem is to use Law of Cosines. For convenience, let the angle formed by $d$ and $n$ be $\alpha$, then $$\cos\alpha =\frac{d^2+n^2-b^2}{2nd}\quad\text{and}\quad\cos(\pi-\alpha)=\frac{d^2+m^2-c^2}{2md}$$ Noting that $\cos\alpha = -\cos(\pi-\alpha)$ yields \begin{align} \frac{d^2+n^2-b^2}{2dn} &= - \frac{d^2+m^2-c^2}{2dm}\\ md^2 + mn^2 - mb^2 &= -(nd^2 +nm^2 - nc^2)\\ (m+n)d^2 + (mn^2+nm^2) &= mb^2 + nc^2\\ ad^2 + mn(m+n) &= b^2m +c^2n\\ a(d^2+mn) &= b^2m+c^2n\\ b^2m + c^2n &=a(d^2+mn) \end{align}