Sunday, February 16, 2014
Triangular angle bisector
Given any triangle with an angle bisector, like:
It can be shown that:
XZXY=ZWWY |
XZsin(180−(α+β))=ZWsinαWYsinα=XYsin(α+β) |
XZsin(180−(α+β))=XZsin(α+β)=ZWsinα∴ |
\begin{aligned} {WY \over \sin{\alpha}} &= {XY \over \sin{(\alpha + \beta)}} \\ \therefore {\sin{(\alpha + \beta)} \over \sin{\alpha}} &= {XY \over WY} \\ \end{aligned} |
\begin{aligned} {XZ \over ZW} &= {XY \over WY} \\ {XZ \over XY} &= {ZW \over WY} \\ \end{aligned} |