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''' | ||
+ | |||
+ | ==Bisectors and tangent== | ||
+ | [[File:Bisectors tangent.png|450px|right]] | ||
+ | Let a triangle <math>\triangle ABC (\angle BAC > \angle BCA)</math> and it’s circumcircle <math>\Omega</math> be given. | ||
+ | |||
+ | 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> | ||
+ | The tangent to <math>\Omega</math> at <math>B</math> meet <math>AC</math> at point <math>M.</math> | ||
+ | Prove that | ||
+ | |||
+ | a)<math>EM = DM = BM,</math> | ||
+ | |||
+ | b)<math> \frac {1}{BM} = \frac {1}{AD} - \frac {1}{CD},</math> | ||
+ | |||
+ | c)<math>\frac {AD^2}{CD^2}=\frac {AM}{CM}.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | 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> | ||
+ | <math>\angle DBE = 90^\circ \implies M</math> is circumcenter <math>\triangle BDE \implies EM = MD.</math> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Proportions for bisectors A== | ||
+ | |||
+ | ==Bisector and circumcircle== | ||
+ | [[File:Bisector divi.png|350px|right]] | ||
+ | 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> | ||
+ | The lines <math>AA', BB',</math> and <math>CC'</math> meet circumcircle <math>ABC (\Omega</math> at points <math>D, E, F,</math> respectively. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | <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>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>\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>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>\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> | ||
+ | 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''' |
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
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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)
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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.
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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
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