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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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| − | ==Bisectors and tangent==
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| − | [[File:Bisectors tangent.png|450px|right]]
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| − | Let a triangle <math>\triangle ABC (\angle BAC > \angle BCA)</math> and it’s circumcircle <math>\Omega</math> be given.
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| − | Let segments <math>BD, D \in AC</math> and <math>BE, E \in AC,</math> be the internal and external bisectors of <math>\triangle ABC.</math>
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| − | The tangent to <math>\Omega</math> at <math>B</math> meet <math>AC</math> at point <math>M.</math>
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| − | Prove that
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| − | a)<math>EM = DM = BM,</math>
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| − | b)<math> \frac {1}{BM} = \frac {1}{AD} - \frac {1}{CD},</math>
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| − | c)<math>\frac {AD^2}{CD^2}=\frac {AM}{CM}.</math>
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| − | <i><b>Proof</b></i>
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| − | a) <math>\angle ABM = \angle ACB = \frac {\overset{\Large\frown} {AB}}{2} \implies \angle ADB = \angle CBD + \angle BCD = \angle ABD + \angle ABM = \angle MBD \implies BM = DM.</math>
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| − | <math>\angle DBE = 90^\circ \implies M</math> is circumcenter <math>\triangle BDE \implies EM = MD.</math>
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| − | b) <math> \frac {AD}{DC} = \frac {AB}{BC} = \frac {AE}{CE} = \frac {DE – AD}{DE + CD} \implies \frac {1}{AD} - \frac {1}{CD} = \frac {2}{DE} =\frac {1}{BM}.</math>
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| − | c) <cmath> \frac {AM}{CM} = \frac {MD – AD}{MD + CD} =\frac {1 –\frac {AD}{BM}}{1 + \frac {CD}{BM}} = \frac {AD}{CD} : \frac {CD}{AD} = \frac {AD^2}{CD^2}.</cmath>
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| − | '''vladimir.shelomovskii@gmail.com, vvsss'''
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| − | ==Proportions for bisectors==
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| − | [[File:Bisector 60.png|400px|right]]
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| − | 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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| − | Prove <math>\frac {CD}{AE} = \frac {BC}{AB}, DI = IE.</math>
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| − | <i><b>Proof</b></i>
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| − | Denote the angles <math>A = 2\alpha, B = 2\beta = 60^\circ, C = 2 \gamma.</math>
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| − | <math>\angle AIC = 180^\circ - \alpha - \gamma = 90^\circ + \beta = 120^\circ \implies B, D, I,</math> and <math>E</math> are concyclic.
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| − | <cmath>\angle BEA = \angle BEI = \angle ADC.</cmath>
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| − | The area of the <math>\triangle ABC</math> is
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| − | <cmath>[ABC] = AB \cdot h_C = AB \cdot CD \cdot \sin \angle ADC = BC \cdot AE \cdot \sin \angle AEB \implies</cmath>
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| − | <cmath>\frac {CD}{AE} = \frac {BC}{AB} = \frac {a}{c}.</cmath>
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| − | <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>
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| − | '''vladimir.shelomovskii@gmail.com, vvsss'''
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| − | ==Bisector and circumcircle==
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| − | [[File:Bisector divi.png|350px|right]]
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| − | Let a triangle <math>\triangle ABC, BC = a, AC = b, AB = c</math> be given.
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| − | Let segments <math>AA', BB',</math> and <math>CC'</math> be the bisectors of <math>\triangle ABC.</math>
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| − | 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 <math>J</math> of <math>\triangle BA'I</math> lies on <math>DF.</math>
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| − | <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>
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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},</cmath>
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| − | <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>
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| − | <cmath>\frac {B'I}{B'E} = \frac {a+c}{b} -1.</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>
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| − | <cmath>AI = \sqrt {bc \frac {b+c-a}{a+b+c}}, FI = c \sqrt {\frac {ab}{(a+b-c)(a+b+c)}}.</cmath>
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| − | <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>
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| − | <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>
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| − | Incenter <math>J</math> belong the bisector <math>BI</math> which is the median of isosceles <math>\triangle IDB.</math>
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| − | '''vladimir.shelomovskii@gmail.com, vvsss'''
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