Difference between revisions of "Bisector"
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Similarly <math>BC' = \frac {a \cdot c}{a+b}, B'C = \frac {a \cdot b}{a+b}. </math> | 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 {BI}{IB'} = \frac {a}{B'C} = \frac{a+c}{b} \implies \frac {BI}{BB'} = \frac {a+c}{a + b +c} \implies \frac {B'I}{BI} = \frac {B'B - BI}{BI} =\frac {b}{a+c}.</cmath> |
<cmath> \frac {DA'}{DC'} = \frac {BA'}{BC'} = \frac {a+ b}{b +c}.</cmath> | <cmath> \frac {DA'}{DC'} = \frac {BA'}{BC'} = \frac {a+ b}{b +c}.</cmath> | ||
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Bisector <math>BD = 2 \frac {BC' \cdot BA'}{BC' + BA'} \cos \beta \implies</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> | + | <cmath>\frac {BD}{BB'} = \frac{a+c}{a+2b+c} \implies \frac {B'D}{BD} = \frac {BB' - BD}{BD} = \frac{2b}{a+c} = 2\frac {B'I}{BI}.</cmath> |
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
==Bisectors and tangent== | ==Bisectors and tangent== | ||
[[File:Bisectors tangent.png|450px|right]] | [[File:Bisectors tangent.png|450px|right]] | ||
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'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
− | ==Proportions for bisectors | + | ==Proportions for bisectors A== |
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==Bisector and circumcircle== | ==Bisector and circumcircle== | ||
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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''' | ||
+ | ==Some properties of the angle bisectors== | ||
+ | [[File:Bisector division B.png|450px|right]] | ||
+ | Let a triangle <math>\triangle ABC, BC = a, AC = b, AB = c,</math> | ||
+ | <math>\angle BAC = 2\alpha, \angle ABC = 2\beta, \angle ACB = 2\gamma</math> be given. | ||
+ | |||
+ | Let <math>R, \Omega, O, r, \omega, I</math> be the circumradius, circumcircle, circumcenter, inradius, incircle, and inradius of <math>\triangle ABC,</math> respectively. | ||
+ | |||
+ | Let segments <math>AA', BB',</math> and <math>CC'</math> be the angle bisectors of <math>\triangle ABC,</math> lines <math>AA', BB',</math> and <math>CC'</math> meet <math>\Omega</math> at <math>D,E,</math> and <math>F, \omega</math> meet <math>BC, AC,</math> and <math>AB</math> at <math>A'', B'', C''.</math> | ||
+ | |||
+ | Let <math>N</math> be the point on tangent to <math>\Omega</math> at point <math>B</math> such, that <math>NI || AC.</math> | ||
+ | |||
+ | Let bisector <math>AB</math> line <math>FM</math> meet <math>BB'</math> at point <math>H</math> and <math>AA'</math> at point <math>G (O \in FM).</math> | ||
+ | |||
+ | Denote <math>Q</math> circumcenter of <math>\triangle ABB', P</math> - the point where bisector <math>AA'</math> meet circumcircle of <math>\triangle ABB'.</math> | ||
+ | |||
+ | Prove:<math> a) BN = \frac {2Rr}{|a-c|},</math> <math>b) \frac {FQ}{QG} = \frac {a}{c},</math> | ||
+ | |||
+ | c) lines <math>FD, A'C',</math> and <math>MP</math> are concurrent at <math>N.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | WLOG, <math>\alpha > \gamma.</math> A few preliminary formulas: | ||
+ | <cmath>\alpha + \beta + \gamma = 90^\circ \implies \sin (\alpha + \beta) = \cos \gamma.</cmath> | ||
+ | <cmath>\frac {a-b}{c} = \frac {\sin 2\alpha - \sin 2 \beta}{\sin 2\gamma} = \frac {2 \sin (\alpha - \beta) \cos(\alpha + \beta)} {2 \sin (\alpha + \beta) \cos (\alpha + \beta)} = \frac {\sin (\alpha - \beta)}{\cos \gamma}.</cmath> | ||
+ | <cmath>\frac {a+b}{c} = \frac {\sin 2\alpha + \sin 2 \beta}{\sin 2\gamma} = \frac {2 \sin (\alpha + \beta) \cos(\alpha - \beta)} {2 \sin (\alpha + \beta) \cos (\alpha + \beta)} = \frac {\cos (\alpha - \beta)}{\sin \gamma}.</cmath> | ||
+ | <cmath>a^2 + c^2 - 2ac\cos 2\beta = b^2 \implies 4 \cos^2 \beta = \frac {(a+b+c)(a+c-b)}{ac}.</cmath> | ||
+ | a) <cmath>\triangle ACC' : \frac {AC}{AC'} = \frac{\sin(180^\circ - 2 \alpha - \gamma)}{\sin \gamma}= \frac{\cos(\alpha - \beta)}{\sin \gamma}= \frac{a+b}{c}.</cmath> | ||
+ | <cmath>\angle FBC' = \gamma, \angle BFC' = 2 \alpha, BF = FI \implies \frac {FI}{FC'} = \frac {a+b}{c}.</cmath> | ||
+ | <cmath>\frac {MG}{MF} = \frac {AM \tan \alpha}{BM \tan \gamma} =\frac {\tan \alpha}{\tan \gamma} = \frac {CB''}{AC''}=\frac{a+b-c}{b+c-a}.</cmath> | ||
+ | <cmath>\angle AOG = 2 \gamma, \angle AGM = 90^\circ - \alpha \implies \angle OAG = |90^\circ - \alpha - 2\gamma| = |\beta - \gamma| \implies \frac {GO}{AO} = \frac {|\sin (\beta - \gamma)|}{\cos \alpha} = \frac{|b-c|}{a}.</cmath> | ||
+ | <cmath>\angle BFD = \alpha,\angle NBD = 2\gamma + 2\beta + \alpha = 180^\circ - \alpha, \angle NBF = \angle BDF = \gamma \implies</cmath> | ||
+ | <cmath>\frac {NB}{NF} = \frac {ND}{NB} = \frac {\sin \gamma}{\sin \alpha} \implies \frac {NF}{ND} = \frac {\sin^2 \gamma}{\sin^2 \alpha} = \frac {\sin 2\gamma}{\sin 2\alpha} \cdot \frac {\tan \gamma}{\tan \alpha} = \frac {c}{a} \cdot \frac{b+c-a}{a+b-c}.</cmath> | ||
+ | <cmath>\angle NBI = \angle NIB = 2\gamma + \beta = 90^\circ +\gamma - \alpha \implies \cos \angle NBI = \sin (\alpha - \gamma).</cmath> | ||
+ | |||
+ | <cmath>BI = \frac {BC''}{\cos \beta} = \frac {a+c-b}{2\cos \beta} \implies NB = \frac {BI}{2 \sin |\alpha - \gamma|} | ||
+ | = \frac{a+c-b}{4\cos^2 \beta} \cdot \frac {\cos \beta}{\sin |\alpha - \gamma|} = \frac{abc}{|a-c|(a+b+c)} = \frac {2Rr}{|a-c|}.</cmath> | ||
+ | |||
+ | b)<cmath>\triangle AIC \sim \triangle FID, k = \frac {IB''}{IL} = \frac {2r}{IB} = 2 \sin \beta \implies FD = \frac {AC}{k} = \frac {b}{2 \sin \beta}.</cmath> | ||
+ | <math>Q</math> is the circumcenter of <math>\triangle ABB' \implies \angle BQM = \angle AB'B \implies \angle ABQ = \alpha - \gamma.</math> | ||
+ | <cmath>BQ = \frac {BM}{\cos (\alpha - \gamma)} = \frac {c}{2} \cdot \frac {b}{(a+c) \sin \beta} = \frac {bc}{2(a+c) \sin \beta}.</cmath> | ||
+ | <cmath>PQ \perp BB' \implies PQ || FD \implies \triangle GQP \sim \triangle GFD, k = \frac{FD}{QP} = \frac{FD}{QB} = \frac {a+c}{c} \implies \frac {FQ}{QG} = k - 1 = \frac {a}{c} = \frac {DP}{PG}.</cmath> | ||
+ | |||
+ | c)<math>BF = FI, BD = DI, BN = NI \implies N, F, D</math> are collinear. | ||
+ | |||
+ | <cmath>\frac {NF}{ND} = \frac {c}{a} \cdot \frac {b+c-a}{a+b-c}, \frac {IC'}{C'F} = \frac {a+b-c}{c}, \frac {IA'}{A'D} = \frac {c+b-a}{a} \implies</cmath> | ||
+ | <math>N, C', A'</math> are collinear and so on. Using Cheva's theorem we get the result. | ||
+ | |||
+ | '''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''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Latest revision as of 09:47, 26 March 2024
Contents
Division of bisector
Let a triangle be given.
Let and be the bisectors of
he segments and meet at point Find
Solution
Similarly
Denote Bisector
Bisector vladimir.shelomovskii@gmail.com, vvsss
Bisectors and tangent
Let a triangle and it’s circumcircle be given.
Let segments and be the internal and external bisectors of The tangent to at meet at point Prove that
a)
b)
c)
Proof
a) is circumcenter
b)
c) vladimir.shelomovskii@gmail.com, vvsss
Proportions for bisectors A
Bisector and circumcircle
Let a triangle be given. Let segments and be the bisectors of The lines and meet circumcircle at points respectively.
Find Prove that circumcenter of lies on
Solution
Incenter belong the bisector which is the median of isosceles
vladimir.shelomovskii@gmail.com, vvsss
Some properties of the angle bisectors
Let a triangle be given.
Let be the circumradius, circumcircle, circumcenter, inradius, incircle, and inradius of respectively.
Let segments and be the angle bisectors of lines and meet at and meet and at
Let be the point on tangent to at point such, that
Let bisector line meet at point and at point
Denote circumcenter of - the point where bisector meet circumcircle of
Prove:
c) lines and are concurrent at
Proof
WLOG, A few preliminary formulas: a)
b) is the circumcenter of
c) are collinear.
are collinear and so on. Using Cheva's theorem we get the result.
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. The area of the is vladimir.shelomovskii@gmail.com, vvsss