proof of Steiner’s theorem

Using $\alpha ,\beta ,\gamma ,\delta$ to denote angles as in the diagram at 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).

Title	proof of Steiner’s theorem
Canonical name	ProofOfSteinersTheorem
Date of creation	2013-03-22 13:48:06
Last modified on	2013-03-22 13:48:06
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	5
Author	mathcam (2727)
Entry type	Proof
Classification	msc 51N20