Geometrical proof that $b\cos\beta+c\cos\gamma=a\cos(\beta-\gamma)$ 
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.
 A: 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.)
A: The answer was triggered by this question.
Consider triangle $ABC$
with side lengths $a,b,c$ and corresponding angles $\alpha,\beta,\gamma$,
circumscribed circle with the center $O$,
line $DE$ tangent to the circle at $A$,
$BD\perp DE$, $CE\perp DE$, $BF\perp CE$:

Then we have
\begin{align}
|DE|&=|BF|
,\\
\triangle ACE,\triangle ABD:\quad
|DE|&=|AE|+|AD|=b\cos\beta+c\cos\gamma
,\\
\triangle BCF:\quad
|BF|&=|BC|\cos\angle FBC
=|BC|\cos(90^\circ-\angle BCF)
\\
&=
a\cos(90^\circ-(90^\circ-\beta+\gamma))
=
a\cos(\beta-\gamma)
.
\end{align}
