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| − | ==Division of bisector==
| + | #REDIRECT [[Bisection]] |
| − | [[File:Bisector division.png|350px|right]] | |
| − | Let a triangle <math>\triangle ABC, BC = a, AC = b, AB = c</math> be given.
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| − | Let <math>AA', BB',</math> and <math>CC'</math> be the bisectors of <math>\triangle ABC.</math>
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| − | he segments <math>BB'</math> and <math>A'C'</math> meet at point <math>D.</math> Find <cmath>\frac {BI}{BB'}, \frac {DA'}{DC'}, \frac {BD}{BB'}.</cmath>
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| − | <i><b>Solution</b></i>
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| − | <cmath>\frac {BA'}{CA'} = \frac {BA}{CA} = \frac {c}{b}, BA' + CA' = BC = a \implies BA' = \frac {a \cdot c}{b+c}.</cmath>
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| − | Similarly <math>BC' = \frac {a \cdot c}{a+b}, B'C = \frac {a \cdot b}{a+b}. </math>
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| − | <cmath>\frac {BI}{IB'} = \frac {a}{B'C} = \frac{a+c}{b} \implies \frac {BI}{BB'} = \frac {a+c}{a + b +c}.</cmath>
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| − | <cmath> \frac {DA'}{DC'} = \frac {BA'}{BC'} = \frac {a+ b}{b +c}.</cmath>
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| − | Denote <math>\angle ABC = 2 \beta.</math>
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| − | Bisector <math>BB' = 2 \frac {a \cdot c}{a + c} \cos \beta.</math>
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| − | Bisector <math>BD = 2 \frac {BC' \cdot BA'}{BC' + BA'} \cos \beta \implies</math>
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| − | <cmath>\frac {BD}{BB'} = \frac{a+c}{a+2b+c}.</cmath>
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| − | '''vladimir.shelomovskii@gmail.com, vvsss'''
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