Difference between revisions of "Bisector"
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==Bisector and circumcircle== | ==Bisector and circumcircle== | ||
− | [[File:Bisector | + | [[File:Bisector divi.png|350px|right]] |
Let a triangle <math>\triangle ABC, BC = a, AC = b, AB = c</math> be given. | 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> | 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. | 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 of <math>\triangle BA'I</math> lies on <math>DF.</math> | + | 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> | <i><b>Solution</b></i> | ||
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<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>\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 | + | <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>\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 | + | <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>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>\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>\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> | <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> | 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''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 17:45, 9 December 2023
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
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
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