Transversal Trigonometry

The following has been posted by Oai Thanh Dào at the CutTheKnotMath facebook page:

Draw lines through $A$ and $C$ parallel to $B_1C_1$ and then the perpendicular ones through $B$ an $C:$

In this particular case, we observe that

$AD=AB\cos (180^{\circ}-\gamma)=-AB\cos\gamma,\\ AG=AC\cos\beta,\\ DG=CE=BC\cos\alpha.$

Thus we see that, in this configuration, $DG=AG-AD,$ i.e.,

$BC\cos\alpha =AC\cos\beta + AB\cos\gamma.$

Other configurations are treated similarly.

Oai Thanh Dào has also supplied an application for the above: