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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<math>\angle BAC = 2\alpha, \angle ABC = 2\beta, \angle ACB = 2\gamma</math> be given. | <math>\angle BAC = 2\alpha, \angle ABC = 2\beta, \angle ACB = 2\gamma</math> be given. | ||
− | Let <math>\Omega, | + | 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 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 <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 | + | 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> | Prove:<math> a) BN = \frac {2Rr}{|a-c|},</math> <math>b) \frac {FQ}{QG} = \frac {a}{c},</math> | ||
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<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 {\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>\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>\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>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>\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}{ | + | 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 AOG = 2 \gamma, \angle AGM = 90^\circ - \alpha \implies \angle | + | <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>\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{ | + | <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>\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|} | <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> | = \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> | ||
− | <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> | + | 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> |
− | <cmath> | + | <math>Q</math> is the circumcenter of <math>\triangle ABB' \implies \angle BQM = \angle AB'B \implies \angle ABQ = \alpha - \gamma.</math> |
− | <cmath>\triangle | + | <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> | ||
+ | |||
+ | 1. 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> | ||
+ | |||
+ | 2. 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]] | ||
+ | 3.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''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Latest revision as of 06:51, 6 June 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
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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
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Seven lines crossing point
Let be the incenter, circumcircle, and the midpoints of sides
of a
Let be the bisectors of a
be the midpoint of
The points and
be such points that
Denote points
Prove that the lines and the tangent to the circumcircle of
at
are concurrent.
Proof
1. Denote
Similarly
is the bisector of
Similarly,
is the bisector of
is the bisector of
Therefore are rhombus.
So triples of points are collinear, lines
It is known that
Similarly,
is the bisector
Similarly,
Denote the crosspoint of the tangent to the circumcircle of
at
and
is the bisector
2. Let us consider the points and
We use Menelaus' Theorem for and line
and get
We use Menelaus' Theorem for
and get that points
and
are collinear.
3.Let us consider the points and
We use Menelaus' Theorem for
and line
and get
Let
be the midpoint
So
We use Menelaus' Theorem for and get that points
and
are collinear.
Similarly points and
are collinear.
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