Equilateral triangle $BCD$ is constructed out of $\triangle ABC$. Let $AA^\prime$ be diameter of $(ABC)$. If $\angle BDA^\prime=\alpha$, $\angle CDA^\prime=\beta$, prove that \[\frac{\sin(B+\alpha)}{\sin(C+\beta)}=\frac{\cos(B-A)}{\cos(C-A)}.\]
I thought of constructing angles that equal to either of $B+\alpha$, $C+\beta$, $|B-A|$, or $|C-A|$, but cannot do anything meaningful.
Hint: let BC=1 be the radius of unit circle.As can be seen in figure, we have:
$\widehat{B}=\overset\frown{CM}$
We draw a line from C parallel with DG, it meets circle$d_1$ at N, so:
$\angle NBR=(GDB=\alpha)=\overset\frown{RN}$
we connect R to M and draw a line from N parallel with KM, it meets circle $d_1$ at O so we have:
$\overset\frown{CO}=\widehat{B}+\alpha$
Projection of O on vertical from C is point P, hence $BP= \sin(\widehat{B}+\alpha)$
Now we draw a line from B parallel with CA, it meets circle $d_1$ at K, so we have:
$\overset\frown{KS}=\angle SBK=\widehat{C}$
The extension of KB meets the circle at V, so we have:
$BK||AC\Rightarrow \angle BVL=\overset \frown{VL} $
We transfer arc VL such that ray BV lays on CS, the result will be arc BR and arc KR is equal to $C-A$. We tranfer arc KR by parallel lines KC and $RC_1$. The projection of $C_1$ on BC is point $B_1$ and we have:
$\cos (C-A) =BB_1$
The area S of triangle BPC is:
$S= \sin(B+\alpha)\cdot \cos(C-A)$
Similarly find $\sin (C+\beta)$ and $\cos (B-A)$, their product must be equal to S.
I wanted to show a method, I may have some errors in my drawing.