Proving Stewart's theorem without trig Stewart's theorem states that in the triangle shown below,
$$ b^2 m + c^2 n = a (d^2 + mn). $$

Is there any good way to prove this without using any trigonometry? Every proof I can find uses the Law of Cosines.

You can do it using the Pythagorean theorem. I'll treat the case in which $\angle (a,b)$ and $\angle(a,c)$ are both acute and $m \gt n$ as in your figure.
Draw the height $h$ of the triangle $abc$ and denote the resulting middle segment on $a$ by $x$. Using the Pythagorean theorem one sees
$$\begin{align*}
b^2 &= h^2 + (n+x)^2 \\
c^2 &= h^2 + (m-x)^2 \\
d^2 &= h^2 + x^2
\end{align*}$$
Therefore
$$\begin{align*}
b^2 m &= h^2m + n^2m + 2mnx+x^2m \\
c^2 n &= h^2n + m^2n - 2mnx + x^2n \\
b^2 m + c^2n &= (n+m)(h^2  + mn + x^2) = a(d^2 + mn)
\end{align*}$$
as we wanted.

If $\angle(a,c)$ is obtuse, the same idea works: Write
$$b^2 = (m+n+x)^2 + h^2, \qquad c^2 = x^2 + h^2, \qquad d^2 = (m+x)^2 + h^2$$
and calculate similarly.
Symmetry and the consideration of some degenerate cases easily lead to a complete proof of Stewart's theorem from what I've written here.
It would be very nice to have a good pictorial proof that makes the identity similarly obvious as Pythagoras, but so far I couldn't come up with a nice figure that would achieve this.

Added: Matthew Stewart (1717–1785) published this theorem as Proposition II on page 2 in his 1746 book Some general theorems of considerable use in the higher parts of mathematics (arxive.org). Unfortunately, Google's scans don't seem to include the figures, but they are easy to reconstruct from the description in the text.
Update:(by @brainjam) I've taken the liberty of adding a partial scan of the figures from another Google scan -
One case of Stewart's Proposition II reads in the above notation:
$$b^2 + c^2 \frac{n}{m} = an + d^2 \frac{a}{m}$$
and it has the following nice corollary:
$$a^2 + b^2 + c^2 = 2d^2 + m^2 + n^2.$$
Here's the title page and the relevant passage (not using trigonometry either). Since Proposition I is used in the proof, I'm including the entire beginning of the text:







