|
|
| (20 intermediate revisions by one other user not shown) |
| Line 1: |
Line 1: |
| − | ==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.
| |
| − | | |
| − | Let <math>AA', BB',</math> and <math>CC'</math> be the bisectors of <math>\triangle ABC.</math>
| |
| − | | |
| − | 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>
| |
| − | | |
| − | <i><b>Solution</b></i>
| |
| − | | |
| − | <cmath>\frac {BA'}{CA'} = \frac {BA}{CA} = \frac {c}{b}, BA' + CA' = BC = a \implies BA' = \frac {a \cdot c}{b+c}.</cmath>
| |
| − | | |
| − | Similarly <math>BC' = \frac {a \cdot c}{a+b}, B'C = \frac {a \cdot b}{a+b}. </math>
| |
| − | <cmath>\frac {BI}{IB'} = \frac {a}{B'C} = \frac{a+c}{b} \implies \frac {BI}{BB'} = \frac {a+c}{a + b +c}.</cmath>
| |
| − | | |
| − | <cmath> \frac {DA'}{DC'} = \frac {BA'}{BC'} = \frac {a+ b}{b +c}.</cmath>
| |
| − | | |
| − | Denote <math>\angle ABC = 2 \beta.</math>
| |
| − | Bisector <math>BB' = 2 \frac {a \cdot c}{a + c} \cos \beta.</math>
| |
| − | | |
| − | Bisector <math>BD = 2 \frac {BC' \cdot BA'}{BC' + BA'} \cos \beta \implies</math>
| |
| − | <cmath>\frac {BD}{BB'} = \frac{a+c}{a+2b+c}.</cmath>
| |
| − | '''vladimir.shelomovskii@gmail.com, vvsss'''
| |
| − | ==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>
| |
| − |
| |
| − | Prove <math>\frac {CD}{AE} = \frac {BC}{AB}, DI = IE.</math>
| |
| − | | |
| − | <i><b>Proof</b></i>
| |
| − | | |
| − | Denote the angles <math>A = 2\alpha, B = 2\beta = 60^\circ, C = 2 \gamma.</math>
| |
| − | <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>
| |
| − | 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</cmath>
| |
| − | <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'''
| |
