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Angle bisector length in a triangle
02-17-2010, 12:31 AM
Post: #1
Angle bisector length in a triangle
Let \(\Delta_{\mathbf{ABC}}\) be a triangle with side lengths \(a,\; b\) and \(c\) corresponding to angles (vertices) \(A,\; B\) and \(C\).
And let \(D\) be the point where the bisector of \(\angle C\) intersects \(\overline{AB}\). We need to find \(l_{C} =\left|\overline{CD}\right|\).
We have \(\displaystyle {(a+b) l_{c}\sin \frac{C}{2}} = 2\left|\Delta_{\mathbf{ABC}} \right|=ab\sin C \quad\). Therefore[float=left]     [/float]
\[\left|\overline{CD}\right|=l_{c} =\frac{ab \sin C}{(a+b) \sin \frac{C}{2}} = \frac{2ab}{a+b} \cos \frac{C}{2} \]
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