To me, it seems that the Pythagorean theorem is a special case of the law of cosines. However, all derivations that I can find seem to use the Pythagorean theorem in the derivation. Are there any simple (preferably geometric) derivations of the law of cosines that don't reference the Pythagorean theorem?
How to Derive the Law of Cosines Without Using the Pythagorean Theorem?
geometrytrianglestrigonometry
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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:
The law of cosines is:
$$c^2 = a^2 + b^2 \;-\; 2\!\cdot\!a\!\cdot\!b\!\cdot\!\cos\theta$$
where $\theta$ is the angle between the sides $a$ and $b$.
Now, when this angle is a right angle ($90^\circ$, or $\frac{\pi}{2}$), its cosine is $0$, so the entire last term is multiplied by $0$ and vanishes, leaving only the usual Pythagorean Theorem for the right triangle with legs $a$ and $b$ and the hypotenuse $c$:
$$c^2 = a^2 + b^2$$
Simple as that.
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Taken from my answer to the question "What is the most elegant proof of the Pythagorean Theorem?" ...