I was reading Stewart's Theorem which states that
Given a triangle with side lengths $a$, $b$, $c$ and a cevian of length $d$ which divides $a$ into two segments $m$ and $n$ as shown in the figure
we must have $$b^2m+c^2n=a(d^2+mn)$$ which is often remembered using the funny mnemonic, "A man and his dad put a bomb in the sink" which gives $$man+dad=bmb+cnc$$
Now, this can be proved quite easily using The Law of Cosines on $\theta$ and $\theta^\prime$ as marked in the figure. You can find the proof here.
But, I was thinking if there is a more elementary geometric argument (without using trignometry and without using a lot of algebraic manipulation) to see this clearly in front of our eyes. I tried giving an attempt, but it seems extremely difficult using only elementary geometry since there is no trivial symmetry immediately observable in this diagram. Also, the terms involved in the formula, i.e., $b^2m$, $c^2n$, $ad^2$ and $amn$, are all of dimension $[L^3]$- so, it seems difficult to even draw areas on sides and geometrically manipulate them.
Any elegant geometric insight or an intuitive idea would be appreciated.
Note: as pointed out in the comments, and as you can see here as well, the Cosine Law can be very well proved using elementary geometry. While that is of course an answer, we all agree that it's cheating at some level. There's another answer as well, but this one's so much algebraic jargon, I would prefer trigonometry instead.
Also, Stewart's book, Some general theorems of considerable use in the higher parts of mathematics seems to contain a completely geometric argument. But the absense of diagrams in Google Scans combined with the length of this solution makes it very difficult to follow. I would also love to see a summarized version of this in more understandable way without loosing the geometric essence.