Difference between revisions of "Bisector"

Revision as of 03:03, 19 December 2023

Division of bisector

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

Bisectors and tangent

Let a triangle $\triangle ABC (\angle BAC > \angle BCA)$ and it’s circumcircle $\Omega$ be given.

Let segments $BD, D \in AC$ and $BE, E \in AC,$ be the internal and external bisectors of $\triangle ABC.$ The tangent to $\Omega$ at $B$ meet $AC$ at point $M.$ Prove that

a) $EM = DM = BM,$

b) $\frac {1}{BM} = \frac {1}{AD} - \frac {1}{CD},$

c) $\frac {AD^2}{CD^2}=\frac {AM}{CM}.$

Proof

a) $\angle ABM = \angle ACB = \frac {\overset{\Large\frown} {AB}}{2} \implies \angle ADB = \angle CBD + \angle BCD = \angle ABD + \angle ABM = \angle MBD \implies BM = DM.$ $\angle DBE = 90^\circ \implies M$ is circumcenter $\triangle BDE \implies EM = MD.$

b) $\frac {AD}{DC} = \frac {AB}{BC} = \frac {AE}{CE} = \frac {DE – AD}{DE + CD} \implies \frac {1}{AD} - \frac {1}{CD} = \frac {2}{DE} =\frac {1}{BM}.$

c) $\frac {AM}{CM} = \frac {MD – AD}{MD + CD} =\frac {1 –\frac {AD}{BM}}{1 + \frac {CD}{BM}} = \frac {AD}{CD} : \frac {CD}{AD} = \frac {AD^2}{CD^2}.$ vladimir.shelomovskii@gmail.com, vvsss

Proportions for bisectors A

Bisector and circumcircle

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

Some properties of the angle bisectors

Let a triangle $\triangle ABC, BC = a, AC = b, AB = c,$ $\angle BAC = 2\alpha, \angle ABC = 2\beta, \angle ACB = 2\gamma$ be given.

Let $R, \Omega, O, r, \omega, I$ be the circumradius, circumcircle, circumcenter, inradius, incircle, and inradius of $\triangle ABC,$ respectively.

Let segments $AA', BB',$ and $CC'$ be the angle bisectors of $\triangle ABC,$ lines $AA', BB',$ and $CC'$ meet $\Omega$ at $D,E,$ and $F, \omega$ meet $BC, AC,$ and $AB$ at $A'', B'', C''.$

Let $N$ be the point on tangent to $\Omega$ at point $B$ such, that $NI || AC.$

Let bisector $AB$ line $FM$ meet $BB'$ at point $H$ and $AA'$ at point $G (O \in FM).$

Denote $Q$ circumcenter of $\triangle ABB', P$ - the point where bisector $AA'$ meet circumcircle of $\triangle ABB'.$

Prove: $a) BN = \frac {2Rr}{|a-c|},$ $b) \frac {FQ}{QG} = \frac {a}{c},$

c) lines $FD, A'C',$ and $MP$ are concurrent at $N.$

Proof

WLOG, $\alpha > \gamma.$ A few preliminary formulas: $\alpha + \beta + \gamma = 90^\circ \implies \sin (\alpha + \beta) = \cos \gamma.$ $\frac {a-b}{c} = \frac {\sin 2\alpha - \sin 2 \beta}{\sin 2\gamma} = \frac {2 \sin (\alpha - \beta) \cos(\alpha + \beta)} {2 \sin (\alpha + \beta) \cos (\alpha + \beta)} = \frac {\sin (\alpha - \beta)}{\cos \gamma}.$ $\frac {a+b}{c} = \frac {\sin 2\alpha + \sin 2 \beta}{\sin 2\gamma} = \frac {2 \sin (\alpha + \beta) \cos(\alpha - \beta)} {2 \sin (\alpha + \beta) \cos (\alpha + \beta)} = \frac {\cos (\alpha - \beta)}{\sin \gamma}.$ $a^2 + c^2 - 2ac\cos 2\beta = b^2 \implies 4 \cos^2 \beta = \frac {(a+b+c)(a+c-b)}{ac}.$ a) $\triangle ACC' : \frac {AC}{AC'} = \frac{\sin(180^\circ - 2 \alpha - \gamma)}{\sin \gamma}= \frac{\cos(\alpha - \beta)}{\sin \gamma}= \frac{a+b}{c}.$ $\angle FBC' = \gamma, \angle BFC' = 2 \alpha, BF = FI \implies \frac {FI}{FC'} = \frac {a+b}{c}.$ $\frac {MG}{MF} = \frac {AM \tan \gamma}{AM \tan \alpha} = \frac {CB''}{AC''}=\frac{a+b-c}{b+c-a}.$ $\angle AOG = 2 \gamma, \angle AGM = 90^\circ - \alpha \implies \angle OAG = |90^\circ - \alpha - 2\gamma| = |\beta - \gamma| \implies \frac {GO}{AO} = \frac {|\sin (\beta - \gamma)|}{\cos \alpha} = \frac{|b-c|}{a}.$ $\angle BFD = \alpha,\angle NBD = 2\gamma + 2\beta + \alpha = 180^\circ - \alpha, \angle NBF = \angle BDF = \gamma \implies$ $\frac {NB}{NF} = \frac {ND}{NB} = \frac {\sin \gamma}{\sin \alpha} \implies \frac {NF}{ND} = \frac {\sin^2 \gamma}{\sin^2 \alpha} = \frac {\sin 2\gamma}{\sin 2\alpha} \cdot \frac {\tan \gamma}{\tan \alpha} = \frac {c}{a} \cdot \frac{a+b-c}{b+c-a}.$ $\angle NBI = \angle NIB = 2\gamma + \beta = 90^\circ +\gamma - \alpha \implies \cos \angle NBI = \sin (\alpha - \gamma).$

$BI = \frac {BC''}{\cos \beta} = \frac {a+c-b}{2\cos \beta} \implies NB = \frac {BI}{2 \sin |\alpha - \gamma|} = \frac{a+c-b}{4\cos^2 \beta} \cdot \frac {\cos \beta}{\sin |\alpha - \gamma|} = \frac{abc}{|a-c|(a+b+c)} = \frac {2Rr}{|a-c|}.$

b) $\triangle AIC \sim \triangle FID, k = \frac {IB''}{IL} = \frac {2r}{IB} = 2 \sin \beta \implies FD = \frac {AC}{k} = \frac {b}{2 \sin \beta}.$ $S$ - circumcenter $\triangle ABB' \implies \angle BSM = \angle AB'B \implies \angle ABS = \alpha - \gamma.$ $BS = \frac {BM}{\cos (\alpha - \gamma)} = \frac {c}{2} \cdot \frac {b}{(a+c) \sin \beta} = \frac {bc}{2(a+c) \sin \beta}.$ $\triangle GSP \sim \triangle GFD, k = \frac{FD}{SP} = \frac{FD}{SB} = \frac {a+c}{c} \implies \frac {FS}{SG} = k - 1 = \frac {a}{c} = \frac {DP}{PG}.$ vladimir.shelomovskii@gmail.com, vvsss

Proportions for bisectors

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

@@ Line 78: / Line 78: @@
 <math>\angle BAC = 2\alpha, \angle ABC = 2\beta, \angle ACB = 2\gamma</math> be given.
-Let <math>\Omega,R,O,\omega,r,I</math> be the circumradius, circumcircle, circumcenter, inradius, incircle, and inradius of <math>\triangle ABC,</math> respectively.
+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> meet <math>BB'</math> at point <math>H</math> and <math>AA'</math> at point <math>G.</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 line <math>AA'</math> meet circumcircle of <math>\triangle ABB'.</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>
@@ Line 97: / Line 98: @@
 <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 \gamma}{AM \tan \alpha} = \frac {CB''}{AC''}=\frac{a+b-c}{b+c-a}.</cmath>
-<cmath>\angle AOG = 2 \gamma, \angle AGM = 90^\circ - \alpha \implies \angle OGA = |90^\circ - \alpha - 2\gamma| = |\beta - \gamma| \implies \frac {GO}{AO} = \frac {|\sin (\beta - \gamma)|}{\cos \alpha} = \frac{|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{a+b-c}{b+c-a}.</cmath>
 <cmath>\angle NBI = \angle NIB = 2\gamma + \beta = 90^\circ +\gamma - \alpha \implies \cos \angle NBI = \sin (\alpha - \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>
 <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>

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Difference between revisions of "Bisector"

Revision as of 03:03, 19 December 2023

Contents

Division of bisector

Bisectors and tangent

Proportions for bisectors A

Bisector and circumcircle

Some properties of the angle bisectors

Proportions for bisectors