Difference between revisions of "Symmetry"

Revision as of 07:15, 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 $AB = AC,$ then $\angle C = \angle B;$ the other cases hold by symmetry because the steps are the same.

Hidden symmetry

Let the convex quadrilateral $ABCD$ be given. $AC = DE, \angle CAD + \angle ACB = 180^\circ.$

Prove that $\angle ABC = \angle ADC.$

Proof

Let $\ell$ be bisector $AC.$

Let point $E$ be symmetric $D$ with respect $\ell.$

$\angle CAD = \angle ACE \implies \angle CAD + \angle ACB = 180^\circ \implies E \in BC.$ $AE = CD = AB \implies \triangle ABE$ is isosceles.

Therefore $\angle ABC = \angle AEC = \angle ADC \blacksquare.$ vladimir.shelomovskii@gmail.com, vvsss

Symmetry with respect angle bisectors

Given the triangle $\triangle ABC, \omega$ is the incircle, $I$ is the incenter, $B' = \omega \cap AC.$

Points $D$ and $E$ are symmetrical to point $B$ with respect to the lines containing the bisectors $AI$ and $CI,$ respectively.

Prove that $B'$ is the midpoint $DE.$

Proof $B \in AB \implies D \in AC, B \in CB \implies E \in AC \implies DE \in AC, D \ne E.$ Denote $A' = \omega \cap BC, C' = \omega \cap AB.$

The tangents from point $B$ to $\omega$ are equal $A'B = C'B.$

Point $B'$ is symmetrical to point $C'$ with respect $AI \implies BC'$ is symmetrical to segment $DB' \implies BC' = DB'.$

Symilarly, $BA' = EB' \implies EB' = DB'. \blacksquare.$ vladimir.shelomovskii@gmail.com, vvsss

Composition of symmetries

Let the inscribed convex hexagon $ABCDEF$ be given, $AB || CF || DE, BC ||AD || EF.$ Prove that $\angle ABC = 120^\circ.$

Proof

Denote $O$ the circumcenter of $ABCDEF,$

$\ell$ the common bisector $AB || CF || DE, m$ the common bisector $BC ||AD || EF,$

$\ell \cap m = O, \alpha$ the smaller angle between lines $\ell$ and $m,$

$S_l$ is the symmetry with respect axis $\ell, S_m$ is the symmetry with respect axis $m.$

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.

$B = S_l(A), C = S_m(B) = S_m(S_l(A)) \implies \overset{\Large\frown} {AC} = 2 \alpha.$ $F = S_l(C), E = S_m(F) = S_m(S_l(C)) \implies \overset{\Large\frown} {CE} = 2 \alpha.$ $D = S_l(E), A = S_m(D) = S_m(S_l(E)) \implies \overset{\Large\frown} {EA} = 2 \alpha.$ Therefore $\overset{\Large\frown} {AC} + \overset{\Large\frown} {CE} + \overset{\Large\frown} {EA} = 6 \alpha = 360^\circ \implies$ $\alpha = 60^\circ \implies \angle ABC = 120^\circ.\blacksquare.$ vladimir.shelomovskii@gmail.com, vvsss

@@ Line 18: / Line 18: @@
 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'''

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Difference between revisions of "Symmetry"

Revision as of 07:15, 29 August 2023

Hidden symmetry

Symmetry with respect angle bisectors

Composition of symmetries