Geometric proof of a trig identity on $\cos t \cos u\cos v$ Consider the following trigonometric identity, valid for any set of angles $u,v,t$: 
$$\cos t⋅\cos u⋅\cos v =\frac14\left[\cos(t + u + v)+\cos(t + u - v)+\cos(u+v-t)+\cos(v + t - u)\right]$$
This identity and its derivation have previously appeared as a question on this site (though the version quoted in the question is in fact missing a term).
But the solutions to the linked question, while valid, were all algebraic in nature.  Is there a geometric proof/demonstration of the above identity? I would also be satisfied with a proof-without-words of a special case e.g. $u=u+v$, $u+v+t=\pi$, $u+v+t=2\pi$, etc.
 A: The key idea is let $P_1$ and $P_2$ be points on a circle $C$ of radius $r$ with center $O$, $P_3:=(P_1+P_2)/2$ be the midpoint, and $P_4$ on circle $C$ be on the ray from center $O$ through point $P_3$. Given a Cartesian coordinate system express
$$ P_1=(r\cos(u+v),r\sin(u+v)), P_2=(r\cos(u-v),r\sin(u-v)),$$
$$P_3=\cos(v)P_4,\; P_4=(r\cos u,r\sin u),.$$
The reason that $P_3=\cos(v)P_4$ is that they are on the same ray from the origin and by rotating the points and/or the coordinate system making $u=0$ gives $P_3=(r\cos(v),0)$ and $P_4=(r,0)$.
The next steps can be done completely geometrically, but I write them algebraically solely for convenience.Thus, setting $r=1$ and using the first coordinates of the points we must have 
$$(\cos(u+v)+\cos(u-v))/2=\cos(v)\cos(u)$$ and similarly for the second coordinates
$$(\sin(u+v)+\sin(u-v))/2=\cos(v)\sin(u).$$
Now apply the first result with $u$ replaced by $u+t$ and then also $u-t$ giving
$$(\cos(u+t+v)+\cos(u+t-v))/2=\cos(v)\cos(u+t)$$
and
$$(\cos(u-t+v)+\cos(u-t-v))/2=\cos(v)\cos(u-t).$$
Averaging the two equations results in the desired trigonometry identity.
