Difference between revisions of "Symmetry"
(→Symmetry with respect angle bisectors 2) |
(→Symmetry with respect angle bisectors 2) |
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==Symmetry with respect angle bisectors 2== | ==Symmetry with respect angle bisectors 2== | ||
[[File:Bisectors 5.png|350px|right]] | [[File:Bisectors 5.png|350px|right]] | ||
+ | [[File:Bisectors 6.png|350px|right]] | ||
Given the triangle <math>\triangle ABC, \omega</math> is the incircle, <math>I</math> is the incenter, | Given the triangle <math>\triangle ABC, \omega</math> is the incircle, <math>I</math> is the incenter, | ||
<cmath>B' = \omega \cap AC, C' = \omega \cap AB.</cmath> | <cmath>B' = \omega \cap AC, C' = \omega \cap AB.</cmath> | ||
Line 71: | Line 72: | ||
<cmath>AB' = AC' \implies \angle AB'C' = \angle AC'B'.</cmath> | <cmath>AB' = AC' \implies \angle AB'C' = \angle AC'B'.</cmath> | ||
− | <math>\angle MB'F = \angle AB'C</math> or <math>\angle MB'F + \angle AB'C = 180^\circ \implies \sin \angle ECM = \sin \angle MB'F.</ | + | <math>\angle MB'F = \angle AB'C</math> or <math>\angle MB'F + \angle AB'C = 180^\circ \implies</math> |
− | <math>\angle EMC' = \angle B'MF</math> or <math>\angle EMC' + \angle B'MF = 180^\circ \implies \sin \angle EMC = \sin \angle B'MF.</ | + | <cmath>\sin \angle ECM = \sin \angle MB'F.</cmath> |
+ | <math>\angle EMC' = \angle B'MF</math> or <math>\angle EMC' + \angle B'MF = 180^\circ \implies</math> | ||
+ | <cmath>\sin \angle EMC' = \sin \angle B'MF.</cmath> | ||
We use the Law of Sines and get: | We use the Law of Sines and get: | ||
− | <cmath>\frac {ME}{\sin {EC'M}} : \frac {MF}{\sin {MB'F}} = \frac {EC'}{\sin {EMC'}} : \frac { | + | <cmath>\frac {ME}{\sin {EC'M}} : \frac {MF}{\sin {MB'F}} = \frac {EC'}{\sin {EMC'}} : \frac {B'F}{\sin {B'MF}} \implies \frac {ME}{MF} = \frac {EC'}{B'F} = 1. \blacksquare</cmath> |
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
Revision as of 14:43, 29 August 2023
A proof utilizes symmetry if the steps to prove one thing is identical to those steps of another. For example, to prove that in triangle ABC with all three sides congruent to each other that all three angles are equal, you only need to prove that if then
the other cases hold by symmetry because the steps are the same.
Contents
Hidden symmetry
Let the convex quadrilateral be given.
Prove that
Proof
Let be bisector
Let point be symmetric
with respect
is isosceles.
Therefore
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Symmetry with respect angle bisectors
Given the triangle is the incircle,
is the incenter,
Points and
are symmetrical to point
with respect to the lines containing the bisectors
and
respectively.
Prove that is the midpoint
Proof
Denote
The tangents from point to
are equal
Point is symmetrical to point
with respect
is symmetrical to segment
Symilarly,
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Symmetry with respect angle bisectors 1
The bisector intersect the incircle
of the triangle
at the point
The point
is symmetric to
with respect to
the point
is symmetric to
with respect to
Prove that
is the bisector of the segment
Proof
The point is symmetric to
with respect to
The point is symmetric to
with respect to
Similarly
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Symmetry with respect angle bisectors 2
Given the triangle is the incircle,
is the incenter,
Let be the point on sideline
Points and
are symmetrical to point
with respect to the lines
and
respectively. The line
contains point
Prove that is the midpoint
Proof
The segment is symmetric to
with respect to
the segment
is symmetric to
with respect to
So
Similarly at midpoint
or
or
We use the Law of Sines and get:
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Composition of symmetries
Let the inscribed convex hexagon be given,
Prove that
Proof
Denote the circumcenter of
the common bisector
the common bisector
the smaller angle between lines
and
is the symmetry with respect axis
is the symmetry with respect axis
It is known that the composition of two axial symmetries with non-parallel axes is a rotation centered at point of intersection of the axes at twice the angle from the axis of the first symmetry to the axis of the second symmetry.
Therefore
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Composition of symmetries 1
Let the triangle be given.
is the incircle,
is the incenter,
is the circumcenter of
The point
is symmetric to
with respect to
is symmetric to
with respect to
is symmetric to
with respect to
Prove: a)
b)
Proof
a) Denote the smaller angle between
and
is the symmetry with respect axis
is the symmetry with respect axis
counterclockwise direction.
clockwise direction.
Therefore is parallel to tangent line for
at point
b) is homothetic to
is the circumcenter of
The center of the homothety lies on the line passing through the circumcenters of the triangles.
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