Difference between revisions of "Bisector"
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<cmath>\frac {BD}{BB'} = \frac{a+c}{a+2b+c}.</cmath> | <cmath>\frac {BD}{BB'} = \frac{a+c}{a+2b+c}.</cmath> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Bisectors and tangent== | ||
+ | [[File:Bisectors tangent.png|350px|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== | ==Proportions for bisectors== | ||
[[File:Bisector 60.png|400px|right]] | [[File:Bisector 60.png|400px|right]] |
Revision as of 16:55, 10 December 2023
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
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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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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