Difference between revisions of "Bisector"
(→Division of bisector) |
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'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Proportions for bisectors== | ==Proportions for bisectors== | ||
| + | [[File:Bisector 60.png|400px|right]] | ||
The bisectors <math>AE</math> and <math>CD</math> of a triangle ABC with <math>\angle B = 60^\circ</math> meet at point <math>I.</math> | The bisectors <math>AE</math> and <math>CD</math> of a triangle ABC with <math>\angle B = 60^\circ</math> meet at point <math>I.</math> | ||
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Denote the angles <math>A = 2\alpha, B = 2\beta = 60^\circ, C = 2 \gamma.</math> | Denote the angles <math>A = 2\alpha, B = 2\beta = 60^\circ, C = 2 \gamma.</math> | ||
| − | <math>\angle | + | <math>\angle AIC = 180^\circ - \alpha - \gamma = 90^\circ + \beta = 120^\circ \implies B, D, I,</math> and <math>E</math> are concyclic. |
<cmath>\angle BEA = \angle BEI = \angle ADC.</cmath> | <cmath>\angle BEA = \angle BEI = \angle ADC.</cmath> | ||
The area of the <math>\triangle ABC</math> is | The area of the <math>\triangle ABC</math> is | ||
| − | <cmath>[ABC] = AB \cdot h_C = AB \cdot CD \cdot \sin \angle ADC = BC \cdot AE \cdot \sin \angle AEB \implies \frac {CD}{AE} = \frac {BC}{AB} = \frac {a}{c}.</cmath> | + | <cmath>[ABC] = AB \cdot h_C = AB \cdot CD \cdot \sin \angle ADC = BC \cdot AE \cdot \sin \angle AEB \implies</cmath> |
| − | <cmath>\frac {DI}{IE} = \frac {DI}{ | + | <cmath>\frac {CD}{AE} = \frac {BC}{AB} = \frac {a}{c}.</cmath> |
| + | <cmath>\frac {DI}{IE} = \frac {DI}{CD} \cdot \frac {AE}{IE}\cdot \frac {CD}{AE}= \frac {c}{a+b+c} \cdot \frac {a+b+c} {a} \cdot \frac {a}{c} = 1.</cmath> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
Revision as of 13:47, 8 December 2023
Division of bisector
Let a triangle 
be given.
Let 
and 
be the bisectors of 
he segments 
and 
meet at point 
Find ![\[\frac {BI}{BB'}, \frac {DA'}{DC'}, \frac {BD}{BB'}.\]](http://latex.artofproblemsolving.com/5/1/4/514f96e06e7dd7f44797855eb8ce5f956663504a.png)
Solution
![\[\frac {BA'}{CA'} = \frac {BA}{CA} = \frac {c}{b}, BA' + CA' = BC = a \implies BA' = \frac {a \cdot c}{b+c}.\]](http://latex.artofproblemsolving.com/4/f/d/4fde431607d6e48344816a4f464a1e111879b731.png)
Similarly 
![\[\frac {BI}{IB'} = \frac {a}{B'C} = \frac{a+c}{b} \implies \frac {BI}{BB'} = \frac {a+c}{a + b +c}.\]](http://latex.artofproblemsolving.com/c/2/9/c29e89101ea63d35bfc04f787c84588fba11f410.png)
![\[\frac {DA'}{DC'} = \frac {BA'}{BC'} = \frac {a+ b}{b +c}.\]](http://latex.artofproblemsolving.com/a/a/4/aa4500b7d5b2f65f09c84e3455da66dccb16ca1d.png)
Denote 
Bisector 
Bisector 
![\[\frac {BD}{BB'} = \frac{a+c}{a+2b+c}.\]](http://latex.artofproblemsolving.com/3/4/e/34edec2d0145ed29a1b4137b26a852d12efdc5e3.png)
vladimir.shelomovskii@gmail.com, vvsss
Proportions for bisectors
The bisectors 
and 
of a triangle ABC with 
meet at point 
Prove 
Proof
Denote the angles 
and 
are concyclic.
![\[\angle BEA = \angle BEI = \angle ADC.\]](http://latex.artofproblemsolving.com/8/1/7/817ef0fe8e248ac346bd9211c3648f658f76cc16.png)
The area of the 
is
![\[[ABC] = AB \cdot h_C = AB \cdot CD \cdot \sin \angle ADC = BC \cdot AE \cdot \sin \angle AEB \implies\]](http://latex.artofproblemsolving.com/3/c/e/3ce543197c8237b1ddad47827bfddc7b52edf837.png)
![\[\frac {CD}{AE} = \frac {BC}{AB} = \frac {a}{c}.\]](http://latex.artofproblemsolving.com/e/8/5/e85f86be8300dd8c7bcfae5619299d71cac11e5b.png)
![\[\frac {DI}{IE} = \frac {DI}{CD} \cdot \frac {AE}{IE}\cdot \frac {CD}{AE}= \frac {c}{a+b+c} \cdot \frac {a+b+c} {a} \cdot \frac {a}{c} = 1.\]](http://latex.artofproblemsolving.com/9/2/2/922f2cf266527e7be6342e6adf29a4441a3b5b7d.png)
vladimir.shelomovskii@gmail.com, vvsss
