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Sunday, February 16, 2014

 

Triangular angle bisector

Given any triangle with an angle bisector, like:

It can be shown that:
  XZXY=ZWWY
Why ? By the Law of Sines,
  XZsin(180(α+β))=ZWsinαWYsinα=XYsin(α+β)
But sin(180(α+β))=sin(α+β). So,
  XZsin(180(α+β))=XZsin(α+β)=ZWsinα
Recall
  \begin{aligned} {WY \over \sin{\alpha}} &= {XY \over \sin{(\alpha + \beta)}} \\ \therefore {\sin{(\alpha + \beta)} \over \sin{\alpha}} &= {XY \over WY} \\ \end{aligned}
Therefore,
  \begin{aligned} {XZ \over ZW} &= {XY \over WY} \\ {XZ \over XY} &= {ZW \over WY} \\ \end{aligned}
\Box


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