Geometric proof of $\cos A + \cos B + \cos C + \cos (A+B+C) = 4 \cos \frac{A+B}{2} \cos \frac{B+C}{2} \cos \frac{A+C}{2}$ 
Prove:
$$\cos A + \cos B + \cos C + \cos (A+B+C) = 4 \cos \frac{A+B}{2} \cos \frac{B+C}{2} \cos \frac{A+C}{2}$$
Geometrically.

I found the algebraic solutions here but I want to figure out how to calculate the above using geometrical arguments of complex numbers/ vectors.

Some hints of geometry I figured out already:

*

*$\frac{A+B}{2}$ is equidistant in angle mangitude from $A$ and $B$, that is $|\frac{A+B}{2} -A| = | \frac{A+B}{2} - B|$, similar results hold for the other angle averages.

*We may think of the above as summing the projections of three unit vectors onto the x-axis: $ \left( \tau(A) + \tau(B) + \tau(C) + \tau( A+B+C)\right) \cdot \hat{i}$, where $\tau(\phi)$ is the unit vector making angle $\phi$ with the x-axis

Related , related


Consider the eight points on the unit circle at angles $±α ± β ± γ$ from $(1, 0)$, shown in blue. We can construct their centroid in two ways.

*

*Pair them vertically, yielding the four red midpoints
$$(\cos (-α + β + γ), 0),  \quad  (\cos (α - β + γ), 0), \\
   (\cos (α + β - γ), 0),  \quad  (\cos (α + β + γ), 0),$$
and then average these.


*Pair them hierarchically, as shown in green.

*

*First, pair the points separated by angles of $2α$. The resulting level-1 midpoints lie on a circle of radius $\cos α$ at angles $±β ± γ$ from $(\cos α, 0)$.

*Then, pair the level-1 midpoints separated by angles of $2β$. The resulting level-2 midpoints lie on a circle of radius $\cos α \cos β$ at angles $±γ$ from $(\cos α \cos β, 0)$.

*Finally, pair the level-2 midpoints separated by an angle of $2γ$. The resulting level-3 midpoint is $(\cos α \cos β \cos γ, 0)$.



Since the two constructions must be equivalent,
$$\begin{multline*}
\frac{\cos (-α + β + γ) + \cos (α - β + γ) + \cos (α + β - γ) + \cos (α + β + γ)}{4} \\
= \cos α \cos β \cos γ,
\end{multline*}$$
which yields the desired identity when $α = \frac{A + B}{2}, β = \frac{B + C}{2}, γ = \frac{A + C}{2}$.
