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 Angle bisector length in a triangle
02-17-2010, 12:31 AM
Post: #1
 elim Moderator Posts: 573 Joined: Feb 2010 Reputation: 0
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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