Difference between revisions of "Bisector"

Latest revision as of 06:51, 6 June 2024

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

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

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

Bisector $BD = 2 \frac {BC' \cdot BA'}{BC' + BA'} \cos \beta \implies$ $\frac {BD}{BB'} = \frac{a+c}{a+2b+c} \implies \frac {B'D}{BD} = \frac {BB' - BD}{BD} = \frac{2b}{a+c} = 2\frac {B'I}{BI}.$ 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) $\triangle ACC' : \frac {AC}{AC'} = \frac{\sin(180^\circ - 2 \alpha - \gamma)}{\sin \gamma}= \frac{\cos(\alpha - \beta)}{\sin \gamma}= \frac{a+b}{c}.$ $\angle FBC' = \gamma, \angle BFC' = 2 \alpha, BF = FI \implies \frac {FI}{FC'} = \frac {a+b}{c}.$ $\frac {MG}{MF} = \frac {AM \tan \alpha}{BM \tan \gamma} =\frac {\tan \alpha}{\tan \gamma} = \frac {CB''}{AC''}=\frac{a+b-c}{b+c-a}.$ $\angle AOG = 2 \gamma, \angle AGM = 90^\circ - \alpha \implies \angle OAG = |90^\circ - \alpha - 2\gamma| = |\beta - \gamma| \implies \frac {GO}{AO} = \frac {|\sin (\beta - \gamma)|}{\cos \alpha} = \frac{|b-c|}{a}.$ $\angle BFD = \alpha,\angle NBD = 2\gamma + 2\beta + \alpha = 180^\circ - \alpha, \angle NBF = \angle BDF = \gamma \implies$ $\frac {NB}{NF} = \frac {ND}{NB} = \frac {\sin \gamma}{\sin \alpha} \implies \frac {NF}{ND} = \frac {\sin^2 \gamma}{\sin^2 \alpha} = \frac {\sin 2\gamma}{\sin 2\alpha} \cdot \frac {\tan \gamma}{\tan \alpha} = \frac {c}{a} \cdot \frac{b+c-a}{a+b-c}.$ $\angle NBI = \angle NIB = 2\gamma + \beta = 90^\circ +\gamma - \alpha \implies \cos \angle NBI = \sin (\alpha - \gamma).$

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

b) $\triangle AIC \sim \triangle FID, k = \frac {IB''}{IL} = \frac {2r}{IB} = 2 \sin \beta \implies FD = \frac {AC}{k} = \frac {b}{2 \sin \beta}.$ $Q$ is the circumcenter of $\triangle ABB' \implies \angle BQM = \angle AB'B \implies \angle ABQ = \alpha - \gamma.$ $BQ = \frac {BM}{\cos (\alpha - \gamma)} = \frac {c}{2} \cdot \frac {b}{(a+c) \sin \beta} = \frac {bc}{2(a+c) \sin \beta}.$ $PQ \perp BB' \implies PQ || FD \implies \triangle GQP \sim \triangle GFD, k = \frac{FD}{QP} = \frac{FD}{QB} = \frac {a+c}{c} \implies \frac {FQ}{QG} = k - 1 = \frac {a}{c} = \frac {DP}{PG}.$

c) $BF = FI, BD = DI, BN = NI \implies N, F, D$ are collinear.

$\frac {NF}{ND} = \frac {c}{a} \cdot \frac {b+c-a}{a+b-c}, \frac {IC'}{C'F} = \frac {a+b-c}{c}, \frac {IA'}{A'D} = \frac {c+b-a}{a} \implies$ $N, C', A'$ are collinear and so on. Using Cheva's theorem we get the result.

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

Seven lines crossing point

Let $I, \Omega, M, M_0$ be the incenter, circumcircle, and the midpoints of sides $BC, AB$ of a $\triangle ABC.$

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

$A' = AA'' \cap \Omega, B' = BB'' \cap \Omega, C' = CC'' \cap \Omega, L$ be the midpoint of $BB''.$

The points $U \in AA''$ and $V \in CC''$ be such points that $L \in UV, UV \perp BB''.$

Denote points $A_0 = B'C' \cap AB, A_1 = B'C' \cap AC,$ $B_0 = BC \cap A'C', B_1 = AB \cap A'C', C_0 = AC \cap A'B', C_1 = BC \cap A'B'.$

Prove that the lines $A'C', UM_0, A''C'', MV, A_0I,$ and the tangent to the circumcircle of $\triangle ABC$ at $B$ are concurrent.

Proof

1. Denote $AB = c, BC = a, AC = b, \angle BAC = 2 \alpha, \angle ABC = 2 \beta, \angle ACB = 2 \gamma.$ $\overset{\Large\frown} {AB'} = \overset{\Large\frown} {CB'} \implies \angle AC'B' = \angle CC'B'.$ Similarly $\angle AB'C' = \angle BB'C', \angle C'AI = \angle C'IA = \alpha + \gamma \implies B'C'$ is the bisector of $AI.$ Similarly, $A'C'$ is the bisector of $BI, A'B'$ is the bisector of $CI.$

Therefore $AA_0IA_1, BB_0IB_1, CC_0IC_1$ are rhombus.

So triples of points $A_0,I,C_1, B_0,I,A_1, C_0,I,B_1$ are collinear, lines $A_0I || AC, B_0I || AB, C_0I || AC.$ $\triangle ABC \sim \triangle A_0IB_1 \sim \triangle IB_0C_1 \sim \triangle A_1IC_0.$ It is known that $\frac {AI}{IA''} = \frac {b+c}{a}, \frac {BI}{IB''} = \frac {a+c}{b} \implies BB_1 : B_1A_0 : A_0A = a : c : b.$

Similarly, $BB_0 : B_0C_1 : C_1C = c : a : b.$

$IC$ is the bisector $\angle A_0IB_1 \implies \frac {A_0C''}{B_1C''} = \frac {AC}{BC} = \frac {b}{a} \implies BB_1 : B_1C'' : C''A_0 : A_0A = a(a + b) : ac : bc : b(a + b).$

Similarly, $BB_0 : B_0A'' : A''C_1 : C_1C = c(c + b) : ac : ab : b(c + b).$

Denote $D$ the crosspoint of the tangent to the circumcircle of $\triangle ABC$ at $B$ and $A_0I.$

$\angle DBI = \angle BCB' = \angle BCA + 2 \overset{\Large\frown} {AB'} = 2 \gamma + \beta = \angle B_1IA_0 + \angle B_1IB = \angle DIB \implies BD = ID.$ $A'C'$ is the bisector $BI \implies D \in A'C'.$

2. Let us consider the points $A'',C'',$ and $D.$ $\frac {BB_1} {B_1A_0} = \frac {a}{c}, \frac {B_0C_1} {BB_0} = \frac {a}{c}.$

We use Menelaus' Theorem for $\triangle BA_0C_1$ and line $DB_1B_0$ and get $\frac {DA_0} {DC_1} = \frac {c^2}{a^2}.$ $\frac {BC''}{C''A_0} = \frac {BB_1+B_1C''}{C''A_0} = \frac {a(a+b+c)}{bc}.$ $\frac {C_1A''}{BA''} = \frac {C_1A''}{BB_0+B_0A''} = \frac {ab}{c(a+b+c)} \implies \frac {BC''}{C''A_0} \cdot \frac {C_1A''}{BA''} = \frac {a^2}{c^2}.$ We use Menelaus' Theorem for $\triangle BA_0C_1$ and get that points $A'',C'',$ and $D$ are collinear.

3.Let us consider the points $U, M_0,$ and $D.$ $\frac {AM_0}{B_1M_0} = \frac {2AM_0}{2BM_0 - 2BB_1} = \frac {c}{c - 2 \frac{ac}{a+b+c}} = \frac{a +b + c}{b+c - a}.$ $\frac {BA_0}{B_1A_0} = \frac {BB_1 + B_1A_0}{B_1A_0} = \frac {a+c}{c},$ $\frac {C_1B_0}{C_1B} = \frac {C_1B_0}{C_1B_0 + BB_0} = \frac {a}{a + c}.$ We use Menelaus' Theorem for $\triangle BB_0B_1$ and line $DA_0B_1$ and get $\frac {DB_0} {DB_1} = \frac {a}{c}.$ $B_1I || BC, \frac {IA'}{A''A'} = \frac{b+c}{a} = \frac{B_1A'}{B_0A'} \implies \frac{DA'}{DB_1} = \frac {ab}{c(b+c-a)}.$ Let $F$ be the midpoint $BI, FA' || LU \implies \frac {A'I}{A'U} = \frac {FI}{FL} = \frac {BI}{BB'' - BI} = \frac {BI}{B''I} = \frac{a+c}{b}.$ $\frac {A'U}{UA} = \frac {A'I - IU}{AI + IU} = \frac {\frac {A'I}{IU} - 1}{\frac {AI}{IA'} \cdot \frac {A'I}{IU}+1} = \frac{ab}{c(a+b+c)}.$ So $\frac {AM_0}{B_1M_0} \cdot \frac {A'U}{UA} \cdot \frac{DA'}{DB_1} = 1.$

We use Menelaus' Theorem for $\triangle AB_1A'$ and get that points $U, M_0,$ and $D$ are collinear.

Similarly points $V, M,$ and $D$ are collinear.

vladimir.shelomovskii@gmail.com, vvsss

@@ Line 12: / Line 12: @@
 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>
@@ Line 20: / Line 20: @@
 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'''
-==Proportions for bisectors==
-[[File:Bisector 60.png|400px|right]]
+==Bisectors and tangent==
-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>
+[[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
-Prove <math>\frac {CD}{AE} = \frac {BC}{AB}, DI = IE.</math>
+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>
-Denote the angles <math>A = 2\alpha, B = 2\beta = 60^\circ, C = 2 \gamma.</math>
+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 AIC =  180^\circ - \alpha - \gamma =  90^\circ + \beta = 120^\circ \implies B, D, I,</math> and <math>E</math> are concyclic.
+<math>\angle DBE = 90^\circ \implies M</math> is circumcenter <math>\triangle BDE \implies EM = MD.</math>
-<cmath>\angle BEA = \angle BEI = \angle ADC.</cmath>
-The area of the <math>\triangle ABC</math> is
+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>
-<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>
+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>
-<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'''
+==Proportions for bisectors A==
 ==Bisector and circumcircle==
@@ Line 62: / Line 72: @@
 <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'''
+==Seven lines crossing point==
+[[File:2024 11 B.png|390px|right]]
+Let <math>I, \Omega, M, M_0</math> be the incenter, circumcircle, and the midpoints of sides <math>BC, AB</math> of a <math>\triangle ABC.</math>
+Let <math>AA'', BB'', CC''</math> be the bisectors of a <math>\triangle ABC.</math>
+<math>A' = AA'' \cap \Omega, B' = BB'' \cap \Omega, C' = CC'' \cap \Omega, L</math> be the midpoint of <math>BB''.</math>
+The points <math>U \in AA''</math> and <math>V \in CC''</math> be such points that <math>L \in UV, UV \perp BB''.</math>
+Denote points <math>A_0 = B'C' \cap AB, A_1 = B'C' \cap AC,</math>
+<cmath>B_0 = BC \cap A'C', B_1 = AB \cap A'C', C_0 = AC \cap A'B', C_1 = BC \cap A'B'.</cmath>
+Prove that the lines <math>A'C', UM_0, A''C'', MV, A_0I,</math> and the tangent to the circumcircle of <math>\triangle ABC</math> at <math>B</math> are concurrent.
+<i><b>Proof</b></i>
+. Denote <math>AB = c, BC = a, AC = b, \angle BAC = 2 \alpha, \angle ABC = 2 \beta, \angle ACB = 2 \gamma.</math>
+<math>\overset{\Large\frown} {AB'} = \overset{\Large\frown} {CB'} \implies \angle AC'B' = \angle CC'B'.</math>
+Similarly <math>\angle AB'C' = \angle BB'C', \angle C'AI = \angle C'IA = \alpha + \gamma \implies B'C'</math> is the bisector of <math>AI.</math>
+Similarly, <math>A'C'</math> is the bisector of <math>BI, A'B'</math> is the bisector of <math>CI.</math>
+Therefore <math>AA_0IA_1, BB_0IB_1, CC_0IC_1</math> are rhombus.
+So triples of points <math>A_0,I,C_1, B_0,I,A_1, C_0,I,B_1</math> are collinear, lines <math>A_0I || AC, B_0I || AB, C_0I || AC.</math>
+<cmath>\triangle ABC \sim \triangle A_0IB_1 \sim \triangle IB_0C_1 \sim \triangle A_1IC_0.</cmath>
+It is known that <math>\frac {AI}{IA''} = \frac {b+c}{a}, \frac {BI}{IB''} = \frac {a+c}{b} \implies BB_1 : B_1A_0 : A_0A = a : c : b.</math>
+Similarly, <math>BB_0 : B_0C_1 : C_1C = c : a : b.</math>
+<math>IC</math> is the bisector <math>\angle A_0IB_1 \implies \frac {A_0C''}{B_1C''} = \frac {AC}{BC} = \frac {b}{a} \implies BB_1 : B_1C'' : C''A_0 : A_0A = a(a + b) : ac : bc : b(a + b).</math>
+Similarly, <math>BB_0 : B_0A'' : A''C_1 : C_1C = c(c + b) : ac : ab : b(c + b).</math>
+Denote <math>D</math> the crosspoint of the tangent to the circumcircle of <math>\triangle ABC</math> at <math>B</math> and <math>A_0I.</math>
+<math>\angle DBI = \angle BCB' =  \angle BCA + 2 \overset{\Large\frown} {AB'} = 2 \gamma + \beta = \angle B_1IA_0 + \angle B_1IB = \angle DIB \implies BD = ID.</math>
+<math>A'C'</math> is the bisector <math>BI \implies D \in A'C'.</math>
+. Let us consider the points <math>A'',C'',</math> and <math>D.</math>
+<cmath>\frac {BB_1} {B_1A_0} = \frac {a}{c}, \frac {B_0C_1} {BB_0} = \frac {a}{c}.</cmath>
+We use Menelaus' Theorem for <math>\triangle BA_0C_1</math> and line <math>DB_1B_0</math> and get <math>\frac {DA_0} {DC_1} = \frac {c^2}{a^2}.</math>
+<cmath>\frac {BC''}{C''A_0} = \frac {BB_1+B_1C''}{C''A_0} = \frac {a(a+b+c)}{bc}.</cmath>
+<cmath>\frac {C_1A''}{BA''} = \frac {C_1A''}{BB_0+B_0A''} = \frac {ab}{c(a+b+c)} \implies \frac {BC''}{C''A_0} \cdot \frac {C_1A''}{BA''} = \frac {a^2}{c^2}.</cmath>
+We use Menelaus' Theorem for <math>\triangle BA_0C_1</math> and get that points <math>A'',C'',</math> and <math>D</math> are collinear.
+[[File:2024 11 C.png|390px|right]]
+.Let us consider the points <math>U, M_0,</math> and <math>D.</math>
+<cmath>\frac {AM_0}{B_1M_0} = \frac {2AM_0}{2BM_0 - 2BB_1} = \frac {c}{c - 2 \frac{ac}{a+b+c}} = \frac{a +b + c}{b+c - a}.</cmath>
+<cmath>\frac {BA_0}{B_1A_0} = \frac {BB_1 + B_1A_0}{B_1A_0} = \frac {a+c}{c},</cmath>
+<cmath>\frac {C_1B_0}{C_1B} = \frac {C_1B_0}{C_1B_0 + BB_0} = \frac {a}{a + c}.</cmath>
+We use Menelaus' Theorem for <math>\triangle BB_0B_1</math> and line <math>DA_0B_1</math> and get <cmath>\frac {DB_0} {DB_1} = \frac {a}{c}.</cmath>
+<cmath>B_1I || BC, \frac {IA'}{A''A'} = \frac{b+c}{a} = \frac{B_1A'}{B_0A'} \implies \frac{DA'}{DB_1} = \frac {ab}{c(b+c-a)}.</cmath>
+Let <math>F</math> be the midpoint <math>BI, FA' || LU \implies \frac {A'I}{A'U} =  \frac {FI}{FL} = \frac {BI}{BB'' - BI} = \frac {BI}{B''I} = \frac{a+c}{b}.</math>
+<math>\frac {A'U}{UA} =  \frac {A'I - IU}{AI + IU} =  \frac {\frac {A'I}{IU} - 1}{\frac {AI}{IA'} \cdot \frac {A'I}{IU}+1} = \frac{ab}{c(a+b+c)}.</math>
+So <math>\frac {AM_0}{B_1M_0} \cdot \frac {A'U}{UA} \cdot \frac{DA'}{DB_1} = 1.</math>
+We use Menelaus' Theorem for <math>\triangle AB_1A'</math> and get that points <math>U, M_0,</math> and <math>D</math> are collinear.
+Similarly points <math>V, M,</math> and <math>D</math> are collinear.
 '''vladimir.shelomovskii@gmail.com, vvsss'''

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Difference between revisions of "Bisector"

Latest revision as of 06:51, 6 June 2024

Contents

Division of bisector

Bisectors and tangent

Proportions for bisectors A

Bisector and circumcircle

Some properties of the angle bisectors

Proportions for bisectors

Seven lines crossing point