proof of Steiner’s theorem

Using $\alpha ,\beta ,\gamma ,\delta$ to denote angles as in the diagramat left, the sines law yields

${\displaystyle \frac{AB}{\sin (\gamma )}}$$=$${\displaystyle \frac{AC}{\sin (\beta )}}$(1)

${\displaystyle \frac{NB}{\sin (\alpha +\delta )}}$$=$${\displaystyle \frac{NA}{\sin (\beta )}}$(2)

${\displaystyle \frac{MC}{\sin (\alpha +\delta )}}$$=$${\displaystyle \frac{MA}{\sin (\gamma )}}$(3)

${\displaystyle \frac{MB}{\sin (\alpha )}}$$=$${\displaystyle \frac{MA}{\sin (\beta )}}$(4)

${\displaystyle \frac{NC}{\sin (\alpha )}}$$=$${\displaystyle \frac{NA}{\sin (\gamma )}}$(5)

Dividing (2) and (3), and (4) by (5):

$$\frac{MA}{NA}\frac{NB}{MC}=\frac{\sin (\gamma )}{\sin (\beta )}=\frac{NA}{MA}\frac{MB}{NC}$$

and therefore

$$\frac{NB\cdot MB}{MC\cdot NC}=\frac{{\sin }^2(\gamma )}{{\sin }^2(\beta )}=\frac{A{B}^2}{A{C}^2}$$

by (1).