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 13:43, 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
[hide]Hidden symmetry
Let the convex quadrilateral be given.
Prove that
Proof
Let be bisector
Let point be symmetric with respect
is isosceles.
Therefore vladimir.shelomovskii@gmail.com, vvsss
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, vladimir.shelomovskii@gmail.com, vvsss
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 vladimir.shelomovskii@gmail.com, vvsss
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: vladimir.shelomovskii@gmail.com, vvsss
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 vladimir.shelomovskii@gmail.com, vvsss
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.
vladimir.shelomovskii@gmail.com, vvsss