A generalization of formula involving bisector in triangles

Question

I am looking for a generalization of formula involving bisector in triangles, as stated in the picture.

Thanks in advance!

Answer 1

Just apply the sine law to both small triangles and note that $\sin \theta = \sin (180^0 - \theta)$. Then,

$\dfrac {AD}{DB} = \dfrac {AC \sin \alpha}{BC \sin \beta}$.

Answer 2

It's my bad in googling. I found it here https://en.wikipedia.org/wiki/Angle_bisector_theorem