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Difference between revisions of "Bisector"

(→‎Proportions for bisectors)
(→‎Bisector and circumcircle)
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==Bisector and circumcircle==
 
==Bisector and circumcircle==
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[[File:Bisector division.png|350px|right]]
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[[File:Bisector divi.png|350px|right]]
 
Let a triangle <math>\triangle ABC, BC = a, AC = b, AB = c</math> be given.
 
Let a triangle <math>\triangle ABC, BC = a, AC = b, AB = c</math> be given.
 
Let segments <math>AA', BB',</math> and <math>CC'</math> be the bisectors of <math>\triangle ABC.</math>  
 
Let segments <math>AA', BB',</math> and <math>CC'</math> be the bisectors of <math>\triangle ABC.</math>  
 
The lines <math>AA', BB',</math> and <math>CC'</math> meet circumcircle <math>ABC (\Omega</math> at points <math>D, E, F,</math> respectively.  
 
The lines <math>AA', BB',</math> and <math>CC'</math> meet circumcircle <math>ABC (\Omega</math> at points <math>D, E, F,</math> respectively.  
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Find <math>\frac {B'I}{B'E}, \frac {DF}{AC}. </math>
+
 
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Prove that circumcenter of <math>\triangle BA'I</math> lies on <math>DF.</math>
+
Find <math>\frac {B'I}{B'E}, \frac {DF}{AC}.</math>
 +
Prove that circumcenter <math>J</math> of <math>\triangle BA'I</math> lies on <math>DF.</math>
  
 
<i><b>Solution</b></i>
 
<i><b>Solution</b></i>
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<cmath>\frac {B'I}{B'E} = \frac {B'I}{BB'} \cdot \frac {BB'^2}{B'E \cdot BB'} = \frac {a+c}{a + b +c} \cdot \frac {BB'^2}{B'A \cdot B'C}.</cmath>
 
<cmath>\frac {B'I}{B'E} = \frac {B'I}{BB'} \cdot \frac {BB'^2}{B'E \cdot BB'} = \frac {a+c}{a + b +c} \cdot \frac {BB'^2}{B'A \cdot B'C}.</cmath>
  
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<cmath>BB'^2 = 4 \cos^2 \beta \frac {a^2 c^2}{(a+c)^2},  4 \cos^2 \beta = \frac {(a+b+c)(a – b +c)}{ac}, B'A \cdot B'C = \frac {ab}{a+c} \cdot {bc}{a+c} = \frac {a b^2 c}{(a+c)^2}</cmath>
+
<cmath>BB'^2 = 4 \cos^2 \beta \frac {a^2 c^2}{(a+c)^2},  4 \cos^2 \beta = \frac {(a+b+c)(a - b +c)}{ac},</cmath>
 +
<cmath>B'A \cdot B'C = \frac {ab}{a+c} \cdot \frac{bc}{a+c} = \frac {a b^2 c}{(a+c)^2} \implies</cmath>
 
<cmath>\frac {B'I}{B'E} = \frac {a+c}{b} -1.</cmath>
 
<cmath>\frac {B'I}{B'E} = \frac {a+c}{b} -1.</cmath>
  
−
<cmath>\angle IAC = \angle DAC = \angle CFD = \angle IFD, \angle FID = \angle AIC \implies \triangle IFD \sim \triangle IAC.</cmath>
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<cmath>\angle IAC = \angle DAC = \angle CFD = \angle IFD, \angle FID = \angle AIC \implies \triangle IFD \sim \triangle IAC \implies \frac {DF}{AC} = \frac {IF}{AI}.</cmath>
−
<cmath>\frac {DF}{AC} = \frac {IF}{AI}.</cmath>
 
 
<cmath>AI = \sqrt {bc \frac {b+c-a}{a+b+c}}, FI = c \sqrt {\frac {ab}{(a+b-c)(a+b+c)}}.</cmath>
 
<cmath>AI = \sqrt {bc \frac {b+c-a}{a+b+c}}, FI = c \sqrt {\frac {ab}{(a+b-c)(a+b+c)}}.</cmath>
 
<cmath>\frac {DF}{AC} = \frac {IF}{AI} = \sqrt {\frac {ac}{(a+b-c)(-a+b+c)}}.</cmath>
 
<cmath>\frac {DF}{AC} = \frac {IF}{AI} = \sqrt {\frac {ac}{(a+b-c)(-a+b+c)}}.</cmath>
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<cmath>\overset{\Large\frown} {BD} +  \overset{\Large\frown} {FA} +  \overset{\Large\frown} {AE}= \angle BAC + \angle ACB + \angle ABC = 180^circ \implies FD \perp BE.</cmath>
+
<cmath>\overset{\Large\frown} {BD} +  \overset{\Large\frown} {FA} +  \overset{\Large\frown} {AE}= \angle BAC + \angle ACB + \angle ABC = 180^\circ \implies FD \perp BE.</cmath>
 
<cmath>2\angle IBD = 2\angle EBD = \overset{\Large\frown} {EC} +  \overset{\Large\frown} {CD} = \overset{\Large\frown} {AE} +  \overset{\Large\frown} {BD} = 2 \angle BID.</cmath>
 
<cmath>2\angle IBD = 2\angle EBD = \overset{\Large\frown} {EC} +  \overset{\Large\frown} {CD} = \overset{\Large\frown} {AE} +  \overset{\Large\frown} {BD} = 2 \angle BID.</cmath>
 
Incenter <math>J</math> belong the bisector <math>BI</math> which is the median of isosceles <math>\triangle IDB.</math>
 
Incenter <math>J</math> belong the bisector <math>BI</math> which is the median of isosceles <math>\triangle IDB.</math>
  
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 16:45, 9 December 2023

Division of bisector

Bisector division.png

Let a triangle $\triangle ABC, BC = a, AC = b, AB = c$ be given.

Let $AA', BB',$ and $CC'$ be the bisectors of $\triangle ABC.$

he segments $BB'$ and $A'C'$ meet at point $D.$ Find \[\frac {BI}{BB'}, \frac {DA'}{DC'}, \frac {BD}{BB'}.\]

Solution

\[\frac {BA'}{CA'} = \frac {BA}{CA} = \frac {c}{b}, BA' + CA' = BC = a \implies BA' = \frac {a \cdot c}{b+c}.\]

Similarly $BC' = \frac {a \cdot c}{a+b}, B'C = \frac {a \cdot b}{a+b}.$ \[\frac {BI}{IB'} = \frac {a}{B'C} = \frac{a+c}{b} \implies \frac {BI}{BB'} = \frac {a+c}{a + b +c}.\]

\[\frac {DA'}{DC'} = \frac {BA'}{BC'} = \frac {a+ b}{b +c}.\]

Denote $\angle ABC = 2 \beta.$ Bisector $BB' = 2 \frac {a \cdot c}{a + c} \cos \beta.$

Bisector $BD = 2 \frac {BC' \cdot BA'}{BC' + BA'} \cos \beta \implies$ \[\frac {BD}{BB'} = \frac{a+c}{a+2b+c}.\] vladimir.shelomovskii@gmail.com, vvsss

Proportions for bisectors

Bisector 60.png

The bisectors $AE$ and $CD$ of a triangle ABC with $\angle B = 60^\circ$ meet at point $I.$

Prove $\frac {CD}{AE} = \frac {BC}{AB}, DI = IE.$

Proof

Denote the angles $A = 2\alpha, B = 2\beta = 60^\circ, C = 2 \gamma.$ $\angle AIC = 180^\circ - \alpha - \gamma = 90^\circ + \beta = 120^\circ \implies B, D, I,$ and $E$ are concyclic. \[\angle BEA = \angle BEI = \angle ADC.\] The area of the $\triangle ABC$ is \[[ABC] = AB \cdot h_C = AB \cdot CD \cdot \sin \angle ADC = BC \cdot AE \cdot \sin \angle AEB \implies\] \[\frac {CD}{AE} = \frac {BC}{AB} = \frac {a}{c}.\] \[\frac {DI}{IE} = \frac {DI}{CD} \cdot \frac {AE}{IE}\cdot \frac {CD}{AE}= \frac {c}{a+b+c} \cdot \frac {a+b+c} {a} \cdot \frac {a}{c} = 1.\] vladimir.shelomovskii@gmail.com, vvsss

Bisector and circumcircle

Bisector divi.png

Let a triangle $\triangle ABC, BC = a, AC = b, AB = c$ be given. Let segments $AA', BB',$ and $CC'$ be the bisectors of $\triangle ABC.$ The lines $AA', BB',$ and $CC'$ meet circumcircle $ABC (\Omega$ at points $D, E, F,$ respectively.

Find $\frac {B'I}{B'E}, \frac {DF}{AC}.$ Prove that circumcenter $J$ of $\triangle BA'I$ lies on $DF.$

Solution

\[\frac {B'I}{B'E} = \frac {B'I}{BB'} \cdot \frac {BB'^2}{B'E \cdot BB'} = \frac {a+c}{a + b +c} \cdot \frac {BB'^2}{B'A \cdot B'C}.\]

\[BB'^2 = 4 \cos^2 \beta \frac {a^2 c^2}{(a+c)^2}, 4 \cos^2 \beta = \frac {(a+b+c)(a - b +c)}{ac},\] \[B'A \cdot B'C = \frac {ab}{a+c} \cdot \frac{bc}{a+c} = \frac {a b^2 c}{(a+c)^2} \implies\] \[\frac {B'I}{B'E} = \frac {a+c}{b} -1.\]

\[\angle IAC = \angle DAC = \angle CFD = \angle IFD, \angle FID = \angle AIC \implies \triangle IFD \sim \triangle IAC \implies \frac {DF}{AC} = \frac {IF}{AI}.\] \[AI = \sqrt {bc \frac {b+c-a}{a+b+c}}, FI = c \sqrt {\frac {ab}{(a+b-c)(a+b+c)}}.\] \[\frac {DF}{AC} = \frac {IF}{AI} = \sqrt {\frac {ac}{(a+b-c)(-a+b+c)}}.\] \[\overset{\Large\frown} {BD} + \overset{\Large\frown} {FA} + \overset{\Large\frown} {AE}= \angle BAC + \angle ACB + \angle ABC = 180^\circ \implies FD \perp BE.\] \[2\angle IBD = 2\angle EBD = \overset{\Large\frown} {EC} + \overset{\Large\frown} {CD} = \overset{\Large\frown} {AE} + \overset{\Large\frown} {BD} = 2 \angle BID.\] Incenter $J$ belong the bisector $BI$ which is the median of isosceles $\triangle IDB.$

vladimir.shelomovskii@gmail.com, vvsss

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