Difference between revisions of "Symmetry"
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Symilarly, <cmath>BA' = EB' \implies EB' = DB'. \blacksquare.</cmath> | Symilarly, <cmath>BA' = EB' \implies EB' = DB'. \blacksquare.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Symmetry with respect angle bisectors 1== | ||
+ | [[File:Bisectors 3.png|350px|right]] | ||
+ | The bisector <math>BI</math> intersect the incircle <math>\omega</math> of the triangle <math>\triangle ABC</math> at the point <math>K, B' = \omega \cap AC.</math> | ||
+ | The point <math>D</math> is symmetric to <math>K</math> with respect to <math>AI,</math> the point <math>E</math> is symmetric to <math>K</math> with respect to <math>CI.</math> | ||
+ | Prove that <math>B'I</math> is the bisector of the segment <math>DE.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | The point <math>A'= \omega \cap BC</math> is symmetric to <math>C' = \omega \cap BA</math> with respect to <math>AI \implies \overset{\Large\frown} {KA'} = \overset{\Large\frown} {KC'}.</math> | ||
+ | |||
+ | The point <math>C'</math> is symmetric to <math>B'</math> with respect to <math>AI \implies \overset{\Large\frown} {KC'} = \overset{\Large\frown} {DB'}.</math> | ||
+ | |||
+ | Similarly <math>\overset{\Large\frown} {KA'} = \overset{\Large\frown} {EB'} \implies \overset{\Large\frown} {DB'} = \overset{\Large\frown} {EB'} \blacksquare.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Construction of triangle== | ||
+ | [[File:Construction tr.png|350px|right]] | ||
+ | Given points <math>D, E,</math> and <math>F</math> at which the segments of the bisectors <math>AI, BI,</math> and <math>CI,</math> respectively intersect the incircle of <math>\triangle ABC</math> centered at <math>I.</math> | ||
+ | |||
+ | Construct the triangle <math>\triangle ABC.</math> | ||
+ | |||
+ | <i><b>Construction</b></i> | ||
+ | |||
+ | We construct the incenter of <math>\triangle ABC</math> as circumcenter of <math>\odot DEF.</math> | ||
+ | |||
+ | If these points are collinear or if <math>\min(\angle DIE, \angle EIF, \angle DIF) \le 90^\circ</math> construction is impossible. | ||
+ | |||
+ | We construct bisectors <math>BEI</math> and <math>CFI.</math> | ||
+ | |||
+ | We construct the points <math>D'</math> and <math>D''</math> symmetrical to point <math>D</math> with respect to <math>FI</math> and <math>EI,</math> respectively. | ||
+ | |||
+ | We construct the bisector <math>D'D''</math> and choose the point <math>G</math> as the point intersection with the circle <math>\odot DEF,</math> closest to the line <math>D'D''.</math> | ||
+ | |||
+ | We construct a tangent to the the circle <math>\odot DEF,</math> at the point <math>G.</math> It intersects the lines <math>EI</math> and <math>FI</math> at points <math>B</math> and <math>C,</math> respectively. | ||
+ | |||
+ | We construct the tangents to <math>\odot DEF</math> which are symmetrical to sideline <math>BC</math> with respect to <math>BI</math> and <math>CI. \blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Symmetry with respect angle bisectors 2== | ||
+ | [[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, | ||
+ | <cmath>B' = \omega \cap AC, C' = \omega \cap AB.</cmath> | ||
+ | |||
+ | Let <math>D</math> be the point on sideline <math>BC.</math> | ||
+ | |||
+ | Points <math>E</math> and <math>F</math> are symmetrical to point <math>D</math> with respect to the lines <math>BI</math> and <math>CI,</math> respectively. The line <math>\ell \perp EF</math> contains point <math>I.</math> | ||
+ | |||
+ | Prove that <math>M = B'C' \cap \ell</math> is the midpoint <math>EF.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | The segment <math>EC'</math> is symmetric to <math>DA'</math> with respect to <math>BI,</math> the segment <math>FB'</math> is symmetric to <math>DA'</math> with respect to <math>CI.</math> So <math>EC' = FB'.</math> | ||
+ | |||
+ | Similarly <math>ID = IE = IF \implies \ell \cap EF</math> at midpoint <math>EF.</math> | ||
+ | |||
+ | <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</math> | ||
+ | <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: | ||
+ | <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''' | ||
+ | ==Symmetry of radical axes== | ||
+ | [[File:Radical axis.png|350px|right]] | ||
+ | Let triangle <math>\triangle ABC</math> be given. The point <math>I</math> and the circle <math>\Omega</math> are the incenter and the circumcircle of <math>\triangle ABC.</math> | ||
+ | |||
+ | Circle <math>\theta</math> centered at <math>A</math> has the radius <math>AI</math> and intersects <math>\Omega</math> at points <math>D</math> and <math>E.</math> Line <math>\ell</math> is the tangent for <math>\theta</math> at the point <math>I.</math> | ||
+ | |||
+ | Prove that line <math>DE</math> is symmetry to the line <math>BC</math> with respect axis <math>\ell.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <cmath>P = AI \cap \Omega \implies PB = PC = PI \implies</cmath> | ||
+ | circle <math>\Theta</math> centered at <math>P</math> contain points <math>B</math> and <math>C</math> and is tangent for <math>\ell</math> and <math>\theta.</math> | ||
+ | |||
+ | <math>DE</math> is the radical axis of <math>\theta</math> and <math>\Omega.</math> <math>BC</math> is the radical axis of <math>\Theta</math> and <math>\Omega.</math> | ||
+ | |||
+ | <math>\ell</math> is the radical axis of <math>\theta</math> and <math>\Theta \implies BC, DE,</math> and <math>\ell</math> are concurrent (at point <math>F, \ell = FI.</math>) | ||
+ | <cmath>\overset{\Large\frown} {AE} + \overset{\Large\frown} {BD} + \overset{\Large\frown} {BP} = 2 \angle DHI, \overset{\Large\frown} {AD} + \overset{\Large\frown} {BD} + \overset{\Large\frown} {CP} = 2 \angle BGI.</cmath> | ||
+ | <cmath>AE = AD \implies \overset{\Large\frown} {AE} = \overset{\Large\frown} {AD}, BP = CP \implies \overset{\Large\frown} {BP} = \overset{\Large\frown} {CP} \implies \angle DHI = \angle BGI \implies \angle HFI = \angle GFI. \blacksquare</cmath> | ||
+ | |||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
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<cmath>\alpha = 60^\circ \implies \angle ABC = 120^\circ.\blacksquare.</cmath> | <cmath>\alpha = 60^\circ \implies \angle ABC = 120^\circ.\blacksquare.</cmath> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | == Composition of symmetries 1== | ||
+ | [[File:Bisectors 4.png|450px|right]] | ||
+ | Let the triangle <math>\triangle ABC</math> be given. | ||
+ | |||
+ | <math>\omega</math> is the incircle, <math>I</math> is the incenter, <math>O</math> is the circumcenter of <math>\triangle ABC.</math> | ||
+ | <cmath>A' = \omega \cap BC, B' = \omega \cap AC, C' = \omega \cap AB.</cmath> | ||
+ | The point <math>A''</math> is symmetric to <math>A'</math> with respect to <math>AI, B''</math> is symmetric to <math>B'</math> with respect to <math>BI, C''</math> is symmetric to <math>C'</math> with respect to <math>CI.</math> | ||
+ | |||
+ | Prove: a)<math> A''C'' || AC;</math> | ||
+ | |||
+ | b) <math>P = AA'' \cap BB'' \cap CC'' \in IO.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | a) Denote <math>\varphi</math> the smaller angle between <math>AI</math> and <math>CI.</math> | ||
+ | |||
+ | <math>S_A</math> is the symmetry with respect axis <math>AI, S_C</math> is the symmetry with respect axis <math>CI.</math> | ||
+ | |||
+ | <math>A' = S_C(B'), A'' = S_A(A') = S_A(S_C(B')) \implies \overset{\Large\frown} {B'A''} = 2 \varphi</math> counterclockwise direction. | ||
+ | |||
+ | <math>C' = S_A(B'), C'' = S_C(C') = S_C(S_A(B')) \implies \overset{\Large\frown} {B'C''} = 2 \varphi</math> clockwise direction. | ||
+ | |||
+ | Therefore <math>\overset{\Large\frown} {B'A''} = \overset{\Large\frown} {C''B'} \implies A''C''</math> is parallel to tangent line for <math>\omega</math> at point <math>B' \implies A''C'' || AC.</math> | ||
+ | |||
+ | b) <math>A''C'' || AC, A''B'' || AB, B''C'' || BC \implies \triangle ABC</math> is homothetic to <math>\triangle A''B''C''.</math> | ||
+ | |||
+ | <math>\omega</math> is the circumcenter of <math>\triangle A''B''C'' \implies</math> | ||
+ | |||
+ | The center of the homothety lies on the line passing through the circumcenters of the triangles. <math>\blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | == Composition of symmetries 2== | ||
+ | Let triangle <math>\triangle ABC</math> be given. The point <math>I</math> and the circle <math>\omega</math> are the incenter and the incircle of <math>\triangle ABC.</math> | ||
+ | |||
+ | Let <math>S_A</math> be the symmetry with respect axis <math>AI, S_B</math> be the symmetry with respect axis <math>BI, S_C</math> the symmetry with respect axis <math>CI.</math> | ||
+ | <cmath>A' = \omega \cap BC, B' = \omega \cap AC, C' = \omega \cap AB.</cmath> | ||
+ | Find the composition of axial symmetries with respect <math>AI, BI,</math> and <math>CI.</math> | ||
+ | |||
+ | <i><b>Solution</b></i> | ||
+ | |||
+ | It is known that the composition of three axial symmetries whose axes intersect at one point <math>I</math> is an axial symmetry whose axis contains the same point <math>I.</math> | ||
+ | |||
+ | Consider the composition of axial symmetries for point <math>B' : S_A(B') = C', S_B(C') = A', S_C(A') = B' \implies</math> | ||
+ | |||
+ | <math>B'</math> is a fixed point of transformation. | ||
+ | |||
+ | This means that the desired axis of symmetry contains points <math>B'</math> and <math>I</math>, this is a straight line <math>B'I. \blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Symmetry and secant== | ||
+ | [[File:Tangent sym1.png|350px|right]] | ||
+ | [[File:Tangent sym.png|350px|right]] | ||
+ | The circle <math>\omega</math> centered at <math>O</math> and the point <math>A</math> be given. Let <math>AB</math> and <math>AB'</math> be the tangents, <math>ACD</math> be the secant (<math>B \in \omega, B' \in \omega, C \in \omega, D \in \omega, AC < AD).</math> | ||
+ | |||
+ | Segment <math>AD</math> intersects segment <math>BB'</math> at point <math>E.</math> Prove that <math>\frac {AD}{AC} = \frac{DE}{CE}.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>AC'D'</math> be symmetric to <math>ACD</math> with respect the line <math>AO.</math> | ||
+ | <cmath>\overset{\Large\frown} {BC} = \overset{\Large\frown} {B'C'}, \overset{\Large\frown} {BD} = \overset{\Large\frown} {B'D'}, M = CD' \cap C'D \implies M \in AO.</cmath> | ||
+ | |||
+ | It is known that <math>M \in BB' \implies BM = B'M.</math> | ||
+ | <cmath>2\angle CBB' = \overset{\Large\frown} {CC'} + \overset{\Large\frown} {B'C'} = \overset{\Large\frown} {CC'} + \overset{\Large\frown} {BC} = 2\angle BDC'.</cmath> | ||
+ | We use symmetry and get <cmath>\angle BMC = \angle B'MC' = \angle BMD \implies \triangle BCM \sim \triangle DBM.</cmath> | ||
+ | It is known that <math>\triangle ABC \sim \triangle ADB \implies</math> | ||
+ | <cmath>\frac {BD^2}{BC^2} = \frac {AD^2}{AB^2} = \frac {AD^2}{AD \cdot AC} = \frac {AD}{AC}.</cmath> | ||
+ | |||
+ | Triangles <math>\triangle BCM</math> and <math>\triangle DBM</math> have common side <math>BM \implies \frac {[DBM]}{[BCM]} = \frac {DE}{CE}.</math> | ||
+ | |||
+ | Similar triangles <math>\triangle BCM</math> and <math>\triangle DBM</math> have the areas ratio <math>\frac {[DBM]}{[BCM]} = \frac {BD^2}{BC^2} = \frac {AD}{AC}.</math> | ||
+ | |||
+ | Therefore <math>\frac {[DBM]}{[BCM]} = \frac {DE}{CE} = \frac {AD}{AC}. \blacksquare</math> | ||
+ | |||
+ | According the Cross-ratio criterion the four points <math>(A,C; E,D)</math> are a harmonic range (on the real projective line). | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Symmetry and incircle== | ||
+ | [[File:Symmetry incercle 2024 17.png|400px|right]] | ||
+ | Let <math>\triangle ABC</math> with incircle <math>\omega = \odot DEF, D \in BC, E \in AC, F\in AB</math> be given. | ||
+ | |||
+ | Point <math>G = BC \cap EF.</math> | ||
+ | |||
+ | 1. Let <math>P</math> be the point in <math>\overset{\Large\frown} {DE}.</math> | ||
+ | |||
+ | Denote <math>Q = GP \cap \omega, K = \odot BQF \cap GF, L = \odot CPE \cap FE.</math> | ||
+ | |||
+ | Prove that <math>\frac {KE} {FL} = \frac {AB}{AC}.</math> | ||
+ | |||
+ | 2. Let <math>K</math> be the point in the segment <math>GF.</math> | ||
+ | |||
+ | Let <math>L</math> be the point in the ray <math>FE,</math> such that <math>\frac {KE} {FL} = \frac {AB}{AC}.</math> | ||
+ | |||
+ | Denote <math>Q = \odot BFK \cap \omega, P = \odot CEL \cap \omega.</math> | ||
+ | |||
+ | Prove that points <math>G, Q,</math> and <math>P</math> are collinear. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <cmath>1. \angle GFQ = \angle GPE \implies \triangle GFQ \sim \triangle GPE \implies \frac {GF}{GP} = \frac {FQ}{PE}.</cmath> | ||
+ | <cmath>\angle GPF = \angle GEQ \implies \triangle GFP \sim \triangle GQE \implies \frac {FP}{QE} = \frac {GP}{GE} \implies \frac {GF}{GE} = \frac {FQ \cdot FP}{PE \cdot QE}.</cmath> | ||
+ | We use Menelaus theorem for a triangle <math>\triangle AEF</math> and a transversal line <math>GBC</math> and get: | ||
+ | <cmath>\frac {AB \cdot GF \cdot CE}{BF \cdot GE \cdot AC} = 1 \implies \frac {AB}{AC}= \frac{GE \cdot BF}{CE \cdot GF}.</cmath> | ||
+ | <cmath>\angle FKQ = \angle FBQ, \angle FEQ = \angle BFQ \implies \triangle KQE \sim \triangle BQF \implies \frac {KE}{BF} = \frac {QE}{QF}.</cmath> | ||
+ | <cmath>\angle PFE = \angle PEC, \angle ELP = \angle ECP \implies \triangle LPF \sim \triangle CPE \implies \frac {LF}{CE} = \frac {FP}{EP}.</cmath> | ||
+ | <cmath>\frac {KE}{LF} = \frac {BF \cdot QE \cdot EP}{QF \cdot CE \cdot FP} = \frac {BF \cdot GE}{CE \cdot GF} = \frac {AB}{AC}.</cmath> | ||
+ | |||
+ | 2. Denote <math>Q' = GP \cap \omega, K' = \odot BQ'F \cap GF.</math> Then | ||
+ | <cmath>\frac {K'E}{LF} = \frac {AB}{AC} \implies K' = K \implies Q = Q' \in GP.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Symmetry and incircle A== | ||
+ | [[File:Symmetry circle 2024 17.png|400px|right]] | ||
+ | Denote <math>\omega = \odot DEF, P \in \omega</math> is the arbitrary point. <math>Q = GP \cap \omega, Q \ne P.</math> | ||
+ | |||
+ | Prove that <math>BQ, CP,</math> and <math>AD</math> are concurrent. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>I</math> is the incenter of <math>\triangle ABC, I' = GI \cap AD.</math> | ||
+ | |||
+ | Let us make the projective transformations mapping circle <math>\omega</math> onto circle and point <math>I'</math> onto center of this circle. | ||
+ | |||
+ | Denote <math>X_0</math> the result of transformation <math>T</math> of point <math>X: T(X) = X_0.</math> | ||
+ | This transformation maps point <math>G</math> to infinity. | ||
+ | |||
+ | Segment <math>AD</math> this transformation maps onto diameter <math>\omega_0,</math> onto <math>A_0D_0.</math> | ||
+ | |||
+ | We use the cross-ratio <math>\frac {BC \cdot GD}{BD \cdot GC}</math> which is fixed, equation <math>\frac{T(GD)}{T(GC)} = 1,</math> the Claim, and get | ||
+ | |||
+ | <math>\frac {B_0C_0}{B_0D_0} = 2,</math> so <math>D_0</math> is the midpoint of <math>B_0C_0 \implies A_0B_0 = A_0C_0.</math> | ||
+ | |||
+ | Point <math>G_0</math> in infinity, so <math>B_0C_0 || Q_0P_0 || F_0E_0.</math> | ||
+ | |||
+ | Lines <math>B_0Q_0 </math> and <math>C_0P_0</math> are crossing at the line of symmetry <math>A_0D_0,</math> therefore lines <math>BQ, CP,</math> and <math>AD</math> are concurrent. | ||
+ | |||
+ | <i><b>Claim</b></i> | ||
+ | [[File:2024 17 B.png|400px|right]] | ||
+ | Let <math>\triangle ABC</math> with incircle <math>\omega = \odot DEF, D \in BC, E \in AC, F\in AB</math> be given. | ||
+ | |||
+ | Point <math>G = BC \cap EF.</math> Prove <math>\frac {BC \cdot GD} {BD \cdot GC}= 2.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | WLOG, <math>AB < AC.</math> Denote <math>B'</math> the point in <math>AC</math> such that <math>AB' = AB.</math> | ||
+ | <cmath>EB' = FB, BB' || EF \implies \frac {B'C }{BC} = \frac {B'E}{BG} \implies</cmath> | ||
+ | <cmath>BG = \frac {BC \cdot B'E}{B'C} = \frac {BC \cdot BD}{AC - AB}.</cmath> | ||
+ | Denote <math>a= BC, b = AC, c = AB \implies BD = \frac {a+c - b}{2}, CD = \frac {a+b - c}{2},</math> | ||
+ | <cmath>BG = \frac {a(a+c - b)}{2(b - c)}, CG = BG + BC = \frac {a(a + b - c)}{2(b - c)}\implies \frac {BC \cdot GD} {BD \cdot CG}= 2.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Symmetry for 60 degrees angle== | ||
+ | [[File:60 symm.png|350px|right]] | ||
+ | [[File:60 symm1.png|350px|right]] | ||
+ | Let an isosceles triangle <math>\triangle ABC (\angle B = \angle C)</math> be given. | ||
+ | |||
+ | Let <math>BE (E \in AC)</math> be the bisector of <math>\angle B.</math> | ||
+ | |||
+ | a) <math>\angle B = \angle C = 40^\circ.</math> Prove that <math>BE + AE = BC.</math> | ||
+ | |||
+ | b) <math>BE + AE = BC.</math> Prove that <math>\angle B = 40^\circ.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | a) One can find successively angles (see diagram). | ||
+ | |||
+ | b) Let <math>\angle B = 4 \alpha \implies \angle C = 4 \alpha, \angle ABE = \angle CBE = 2 \alpha.</math> | ||
+ | |||
+ | Let <math>BE = BD, D \in BC \implies BC - BE = DC =AE.</math> | ||
+ | |||
+ | Let <math>EF||BC \implies \angle FEB = 2 \alpha = \angle FBE \implies EC = BF = FE.</math> | ||
+ | |||
+ | <cmath>DC = AE, CE = FE, \angle AEF = \angle DCE \implies \triangle AEF = \triangle DCE \implies</cmath> | ||
+ | <cmath>ED = CD \implies \angle CED = 4 \alpha.</cmath> | ||
+ | <cmath>\angle BDE = 90^\circ - \alpha = \angle CED + \angle DCE = 8 \alpha \implies \alpha = 10 ^\circ \implies \angle B = 40^\circ.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==See also== | ||
+ | *[[Symmedian]] | ||
+ | *[[Antiparallel]] |
Latest revision as of 12:44, 6 October 2024
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]- 1 Hidden symmetry
- 2 Symmetry with respect angle bisectors
- 3 Symmetry with respect angle bisectors 1
- 4 Construction of triangle
- 5 Symmetry with respect angle bisectors 2
- 6 Symmetry of radical axes
- 7 Composition of symmetries
- 8 Composition of symmetries 1
- 9 Composition of symmetries 2
- 10 Symmetry and secant
- 11 Symmetry and incircle
- 12 Symmetry and incircle A
- 13 Symmetry for 60 degrees angle
- 14 See also
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
Construction of triangle
Given points and at which the segments of the bisectors and respectively intersect the incircle of centered at
Construct the triangle
Construction
We construct the incenter of as circumcenter of
If these points are collinear or if construction is impossible.
We construct bisectors and
We construct the points and symmetrical to point with respect to and respectively.
We construct the bisector and choose the point as the point intersection with the circle closest to the line
We construct a tangent to the the circle at the point It intersects the lines and at points and respectively.
We construct the tangents to which are symmetrical to sideline with respect to and
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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: vladimir.shelomovskii@gmail.com, vvsss
Symmetry of radical axes
Let triangle be given. The point and the circle are the incenter and the circumcircle of
Circle centered at has the radius and intersects at points and Line is the tangent for at the point
Prove that line is symmetry to the line with respect axis
Proof
circle centered at contain points and and is tangent for and
is the radical axis of and is the radical axis of and
is the radical axis of and and are concurrent (at point )
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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 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.
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Composition of symmetries 2
Let triangle be given. The point and the circle are the incenter and the incircle of
Let be the symmetry with respect axis be the symmetry with respect axis the symmetry with respect axis Find the composition of axial symmetries with respect and
Solution
It is known that the composition of three axial symmetries whose axes intersect at one point is an axial symmetry whose axis contains the same point
Consider the composition of axial symmetries for point
is a fixed point of transformation.
This means that the desired axis of symmetry contains points and , this is a straight line
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Symmetry and secant
The circle centered at and the point be given. Let and be the tangents, be the secant (
Segment intersects segment at point Prove that
Proof
Let be symmetric to with respect the line
It is known that We use symmetry and get It is known that
Triangles and have common side
Similar triangles and have the areas ratio
Therefore
According the Cross-ratio criterion the four points are a harmonic range (on the real projective line).
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Symmetry and incircle
Let with incircle be given.
Point
1. Let be the point in
Denote
Prove that
2. Let be the point in the segment
Let be the point in the ray such that
Denote
Prove that points and are collinear.
Proof
We use Menelaus theorem for a triangle and a transversal line and get:
2. Denote Then vladimir.shelomovskii@gmail.com, vvsss
Symmetry and incircle A
Denote is the arbitrary point.
Prove that and are concurrent.
Proof
Denote is the incenter of
Let us make the projective transformations mapping circle onto circle and point onto center of this circle.
Denote the result of transformation of point This transformation maps point to infinity.
Segment this transformation maps onto diameter onto
We use the cross-ratio which is fixed, equation the Claim, and get
so is the midpoint of
Point in infinity, so
Lines and are crossing at the line of symmetry therefore lines and are concurrent.
Claim
Let with incircle be given.
Point Prove
Proof
WLOG, Denote the point in such that Denote vladimir.shelomovskii@gmail.com, vvsss
Symmetry for 60 degrees angle
Let an isosceles triangle be given.
Let be the bisector of
a) Prove that
b) Prove that
Proof
a) One can find successively angles (see diagram).
b) Let
Let
Let
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