Prove that for triangle $ABC$ with side lengths $a,b,c$ and corresponding angles $\alpha,\beta,\gamma$
\begin{align}b\cos\beta+c\cos\gamma&=a\cos(\beta-\gamma)\tag{1}\label{1} \end{align}
It is straightforward to prove \eqref{1}
by expanding $\cos(\beta-\gamma)$ and expressing $\cos\beta,\sin\beta,\cos\gamma,\sin\gamma$
in terms of $a,b,c$ using the cosine rule.
Both sides of equation \eqref{1} are equal to \eqref{2}:
\begin{align}
\frac{a^2(b^2+c^2)-(b^2-c^2)^2}{2abc}
\tag{2}\label{2}
.
\end{align}
The question is: is there any geometrical proof for \eqref{1}?
One possible geometric construction is shown below.
Best Answer
This is essentially an enhanced comment for @g.kov's answer. The circumcircle and "central" elements are unnecessary clutter. Exchanging $F$ for the fourth corner of the circumscribed rectangle would avoid overlapping elements; moreover, this would make the required angle algebra much simpler. So, a "better" form of the figure is this:
(Of course, the figure assumes (without loss of generality) $\beta\geq \gamma$. It also assumes that $\beta$ is non-obtuse, but it readily adapts to the obtuse case.)