Difference between revisions of "Symmetry"
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Therefore <cmath>\angle ABC = \angle AEC = \angle ADC \blacksquare.</cmath> | Therefore <cmath>\angle ABC = \angle AEC = \angle ADC \blacksquare.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Symmetry with respect angle bisectors== | ||
+ | [[File:Bisectors 1.png|350px|right]] | ||
+ | Given the triangle <math>\triangle ABC, \omega</math> is the incircle, <math>I</math> is the incenter, <math>B' = \omega \cap AC.</math> | ||
+ | |||
+ | Points <math>D</math> and <math>E</math> are symmetrical to point <math>B</math> with respect to the lines containing the bisectors <math>AI</math> and <math>CI,</math> respectively. | ||
+ | |||
+ | Prove that <math>B'</math> is the midpoint <math>DE.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | <cmath>B \in AB \implies D \in AC, B \in CB \implies E \in AC \implies DE \in AC, D \ne E.</cmath> | ||
+ | Denote <math>A' = \omega \cap BC, C' = \omega \cap AB.</math> | ||
+ | |||
+ | The tangents from point <math>B</math> to <math>\omega</math> are equal <math>A'B = C'B.</math> | ||
+ | |||
+ | Point <math>B'</math> is symmetrical to point <math>C'</math> with respect <math>AI \implies BC'</math> is symmetrical to segment <math>DB' \implies BC' = DB'.</math> | ||
+ | |||
+ | Symilarly, <cmath>BA' = EB' \implies EB' = DB'. \blacksquare.</cmath> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
Revision as of 06:15, 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.
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
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