Prove a trigonometric identity: $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$ when $A+B+C=\pi$ There is a trigonometric identity:

$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\equiv 1\text{ when }A+B+C=\pi$$

It is easy to prove it in an algebraic way, just like that:

$\quad\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\\=\cos^2A+\cos^2B+\cos^2\left(\pi-A-B\right)+2\cos A\cos B\cos \left(\pi-A-B\right)\\=\cos^2A+\cos^2B+\cos^2\left(A+B\right)-2\cos A\cos B\cos \left(A+B\right)\\=\cos^2A+\cos^2B+\left(\cos A\cos B-\sin A\sin B\right)^2-2\cos A\cos B\left(\cos A\cos B-\sin A\sin B\right)\\=\cos^2A+\cos^2B+\cos^2A\cos^2B+\sin^2A\sin^2B-2\sin A\cos A\sin B\cos B-2\cos^2A\cos^2B+2\sin A\cos A\sin B\cos B\\=\cos^2A+\cos^2B-\cos^2A\cos^2B+\left(1-\cos^2A\right)\left(1-\cos^2B\right)\\=\cos^2A+\cos^2B-\cos^2A\cos^2B+1-\cos^2A-\cos^2B+\cos^2A\cos^2B\\=1$

Then, I want to find a geometric way to prove this identity, as $A+B+C=\pi$ and it makes me think of the angle sum of triangle. However, it is quite hard to prove it in a geometric way. Therefore, I hope there is someone who can help. Thank you!

For positives $A$, $B$ and $C$ there is the following way.
Let $A=\max\{A,B,C\},$ $\pi-A=\alpha,$ $\frac{\pi}{2}-B=\beta$ and $\frac{\pi}{2}-C=\gamma$.
Thus, $\alpha$, $\beta$ and $\gamma$ are measured angles of the triangle and let sides-lengths of the triangle be $a$, $b$ and $c$ respectively.
Thus, since by law of sines $$\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma},$$ by law of cosine we obtain:
$$\sin^2\alpha=\sin^2\beta+\sin^2\gamma-2\sin\beta\sin\gamma\cos\alpha$$ or
$$\sin^2(\pi-A)=\sin^2\left(\frac{\pi}{2}-B\right)+\sin^2\left(\frac{\pi}{2}-C\right)-2\sin\left(\frac{\pi}{2}-B\right)\sin\left(\frac{\pi}{2}-C\right)\cos(\pi-A)$$ or
$$1-\cos^2A=\cos^2B+\cos^2C+2\cos B\cos C\cos A$$ and we are done!
