# proof of Steiner’s theorem Using $\alpha,\beta,\gamma,\delta$ to denote angles as in the diagram at left, the sines law yields

 $\displaystyle\frac{AB}{\sin(\gamma)}$ $\displaystyle=$ $\displaystyle\frac{AC}{\sin(\beta)}$ (1) $\displaystyle\frac{NB}{\sin(\alpha+\delta)}$ $\displaystyle=$ $\displaystyle\frac{NA}{\sin(\beta)}$ (2) $\displaystyle\frac{MC}{\sin(\alpha+\delta)}$ $\displaystyle=$ $\displaystyle\frac{MA}{\sin(\gamma)}$ (3) $\displaystyle\frac{MB}{\sin(\alpha)}$ $\displaystyle=$ $\displaystyle\frac{MA}{\sin(\beta)}$ (4) $\displaystyle\frac{NC}{\sin(\alpha)}$ $\displaystyle=$ $\displaystyle\frac{NA}{\sin(\gamma)}$ (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

 $\frac{NB\cdot MB}{MC\cdot NC}=\frac{\sin^{2}(\gamma)}{\sin^{2}(\beta)}=\frac{% AB^{2}}{AC^{2}}$

by (1).

Title proof of Steiner’s theorem ProofOfSteinersTheorem 2013-03-22 13:48:06 2013-03-22 13:48:06 mathcam (2727) mathcam (2727) 5 mathcam (2727) Proof msc 51N20