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Difference between revisions of "Symmetry"

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'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''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>
 +
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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.
 +
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Denote <math>X_0</math> the result of transformation <math>T</math> of point <math>X: T(X) = X_0.</math>
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This transformation maps point <math>G</math> to infinity.
 +
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Segment <math>AD</math> this transformation maps onto diameter <math>\omega_0,</math> onto <math>A_0D_0.</math>
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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
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<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>
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Point <math>G_0</math> in infinity, so <math>B_0C_0 || Q_0P_0 || F_0E_0.</math>
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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.
 +
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<i><b>Claim</b></i>
 +
[[File:2024 17 B.png|400px|right]]
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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>
 +
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WLOG, <math>AB < AC.</math> Denote <math>B'</math> the point in <math>AC</math> such that <math>AB' = AB.</math>
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<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>
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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 $AB = AC,$ then $\angle C = \angle B;$ the other cases hold by symmetry because the steps are the same.

Hidden symmetry

Hidden S.png
Hidden Sy.png

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

Bisectors 1.png
Bisectors 2.png

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

Symmetry with respect angle bisectors 1

Bisectors 3.png

The bisector $BI$ intersect the incircle $\omega$ of the triangle $\triangle ABC$ at the point $K, B' = \omega \cap AC.$ The point $D$ is symmetric to $K$ with respect to $AI,$ the point $E$ is symmetric to $K$ with respect to $CI.$ Prove that $B'I$ is the bisector of the segment $DE.$

Proof

The point $A'= \omega \cap BC$ is symmetric to $C' = \omega \cap BA$ with respect to $AI \implies \overset{\Large\frown} {KA'} = \overset{\Large\frown} {KC'}.$

The point $C'$ is symmetric to $B'$ with respect to $AI \implies \overset{\Large\frown} {KC'} = \overset{\Large\frown} {DB'}.$

Similarly $\overset{\Large\frown} {KA'} = \overset{\Large\frown} {EB'} \implies \overset{\Large\frown} {DB'} = \overset{\Large\frown} {EB'} \blacksquare.$

vladimir.shelomovskii@gmail.com, vvsss

Construction of triangle

Construction tr.png

Given points $D, E,$ and $F$ at which the segments of the bisectors $AI, BI,$ and $CI,$ respectively intersect the incircle of $\triangle ABC$ centered at $I.$

Construct the triangle $\triangle ABC.$

Construction

We construct the incenter of $\triangle ABC$ as circumcenter of $\odot DEF.$

If these points are collinear or if $\min(\angle DIE, \angle EIF, \angle DIF) \le 90^\circ$ construction is impossible.

We construct bisectors $BEI$ and $CFI.$

We construct the points $D'$ and $D''$ symmetrical to point $D$ with respect to $FI$ and $EI,$ respectively.

We construct the bisector $D'D''$ and choose the point $G$ as the point intersection with the circle $\odot DEF,$ closest to the line $D'D''.$

We construct a tangent to the the circle $\odot DEF,$ at the point $G.$ It intersects the lines $EI$ and $FI$ at points $B$ and $C,$ respectively.

We construct the tangents to $\odot DEF$ which are symmetrical to sideline $BC$ with respect to $BI$ and $CI. \blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

Symmetry with respect angle bisectors 2

Bisectors 5.png
Bisectors 6.png

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

Let $D$ be the point on sideline $BC.$

Points $E$ and $F$ are symmetrical to point $D$ with respect to the lines $BI$ and $CI,$ respectively. The line $\ell \perp EF$ contains point $I.$

Prove that $M = B'C' \cap \ell$ is the midpoint $EF.$

Proof

The segment $EC'$ is symmetric to $DA'$ with respect to $BI,$ the segment $FB'$ is symmetric to $DA'$ with respect to $CI.$ So $EC' = FB'.$

Similarly $ID = IE = IF \implies \ell \cap EF$ at midpoint $EF.$

\[AB' = AC' \implies \angle AB'C' = \angle AC'B'.\] $\angle MB'F = \angle AB'C$ or $\angle MB'F + \angle AB'C = 180^\circ \implies$ \[\sin \angle ECM = \sin \angle MB'F.\] $\angle EMC' = \angle B'MF$ or $\angle EMC' + \angle B'MF = 180^\circ \implies$ \[\sin \angle EMC' = \sin \angle B'MF.\] We use the Law of Sines and get: \[\frac {ME}{\sin {EC'M}} : \frac {MF}{\sin {MB'F}} = \frac {EC'}{\sin {EMC'}} : \frac {B'F}{\sin {B'MF}} \implies \frac {ME}{MF} = \frac {EC'}{B'F} = 1. \blacksquare\] vladimir.shelomovskii@gmail.com, vvsss

Symmetry of radical axes

Radical axis.png

Let triangle $\triangle ABC$ be given. The point $I$ and the circle $\Omega$ are the incenter and the circumcircle of $\triangle ABC.$

Circle $\theta$ centered at $A$ has the radius $AI$ and intersects $\Omega$ at points $D$ and $E.$ Line $\ell$ is the tangent for $\theta$ at the point $I.$

Prove that line $DE$ is symmetry to the line $BC$ with respect axis $\ell.$

Proof

\[P = AI \cap \Omega \implies PB = PC = PI \implies\] circle $\Theta$ centered at $P$ contain points $B$ and $C$ and is tangent for $\ell$ and $\theta.$

$DE$ is the radical axis of $\theta$ and $\Omega.$ $BC$ is the radical axis of $\Theta$ and $\Omega.$

$\ell$ is the radical axis of $\theta$ and $\Theta \implies BC, DE,$ and $\ell$ are concurrent (at point $F, \ell = FI.$) \[\overset{\Large\frown} {AE} + \overset{\Large\frown} {BD} + \overset{\Large\frown} {BP} = 2 \angle DHI, \overset{\Large\frown} {AD} + \overset{\Large\frown} {BD} + \overset{\Large\frown} {CP} = 2 \angle BGI.\] \[AE = AD \implies \overset{\Large\frown} {AE} = \overset{\Large\frown} {AD}, BP = CP \implies \overset{\Large\frown} {BP} = \overset{\Large\frown} {CP} \implies \angle DHI = \angle BGI \implies \angle HFI = \angle GFI. \blacksquare\]

vladimir.shelomovskii@gmail.com, vvsss

Composition of symmetries

Combination S.png
Combination Sy.png

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

Composition of symmetries 1

Bisectors 4.png

Let the triangle $\triangle ABC$ be given.

$\omega$ is the incircle, $I$ is the incenter, $O$ is the circumcenter of $\triangle ABC.$ \[A' = \omega \cap BC, B' = \omega \cap AC, C' = \omega \cap AB.\] The point $A''$ is symmetric to $A'$ with respect to $AI, B''$ is symmetric to $B'$ with respect to $BI, C''$ is symmetric to $C'$ with respect to $CI.$

Prove: a)$A''C'' || AC;$

b) $P = AA'' \cap BB'' \cap CC'' \in IO.$

Proof

a) Denote $\varphi$ the smaller angle between $AI$ and $CI.$

$S_A$ is the symmetry with respect axis $AI, S_C$ is the symmetry with respect axis $CI.$

$A' = S_C(B'), A'' = S_A(A') = S_A(S_C(B')) \implies \overset{\Large\frown} {B'A''} = 2 \varphi$ counterclockwise direction.

$C' = S_A(B'), C'' = S_C(C') = S_C(S_A(B')) \implies \overset{\Large\frown} {B'C''} = 2 \varphi$ clockwise direction.

Therefore $\overset{\Large\frown} {B'A''} = \overset{\Large\frown} {C''B'} \implies A''C''$ is parallel to tangent line for $\omega$ at point $B' \implies A''C'' || AC.$

b) $A''C'' || AC, A''B'' || AB, B''C'' || BC \implies \triangle ABC$ is homothetic to $\triangle A''B''C''.$

$\omega$ is the circumcenter of $\triangle A''B''C'' \implies$

The center of the homothety lies on the line passing through the circumcenters of the triangles. $\blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

Composition of symmetries 2

Let triangle $\triangle ABC$ be given. The point $I$ and the circle $\omega$ are the incenter and the incircle of $\triangle ABC.$

Let $S_A$ be the symmetry with respect axis $AI, S_B$ be the symmetry with respect axis $BI, S_C$ the symmetry with respect axis $CI.$ \[A' = \omega \cap BC, B' = \omega \cap AC, C' = \omega \cap AB.\] Find the composition of axial symmetries with respect $AI, BI,$ and $CI.$

Solution

It is known that the composition of three axial symmetries whose axes intersect at one point $I$ is an axial symmetry whose axis contains the same point $I.$

Consider the composition of axial symmetries for point $B' : S_A(B') = C', S_B(C') = A', S_C(A') = B' \implies$

$B'$ is a fixed point of transformation.

This means that the desired axis of symmetry contains points $B'$ and $I$, this is a straight line $B'I. \blacksquare$

vladimir.shelomovskii@gmail.com, vvsss

Symmetry and secant

Tangent sym1.png
Tangent sym.png

The circle $\omega$ centered at $O$ and the point $A$ be given. Let $AB$ and $AB'$ be the tangents, $ACD$ be the secant ($B \in \omega, B' \in \omega, C \in \omega, D \in \omega, AC < AD).$

Segment $AD$ intersects segment $BB'$ at point $E.$ Prove that $\frac {AD}{AC} = \frac{DE}{CE}.$

Proof

Let $AC'D'$ be symmetric to $ACD$ with respect the line $AO.$ \[\overset{\Large\frown} {BC} = \overset{\Large\frown} {B'C'}, \overset{\Large\frown} {BD} = \overset{\Large\frown} {B'D'}, M = CD' \cap C'D \implies M \in AO.\]

It is known that $M \in BB' \implies BM = B'M.$ \[2\angle CBB' = \overset{\Large\frown} {CC'} + \overset{\Large\frown} {B'C'} = \overset{\Large\frown} {CC'} + \overset{\Large\frown} {BC} = 2\angle BDC'.\] We use symmetry and get \[\angle BMC = \angle B'MC' = \angle BMD \implies \triangle BCM \sim \triangle DBM.\] It is known that $\triangle ABC \sim \triangle ADB \implies$ \[\frac {BD^2}{BC^2} = \frac {AD^2}{AB^2} = \frac {AD^2}{AD \cdot AC} = \frac {AD}{AC}.\]

Triangles $\triangle BCM$ and $\triangle DBM$ have common side $BM \implies \frac {[DBM]}{[BCM]} = \frac {DE}{CE}.$

Similar triangles $\triangle BCM$ and $\triangle DBM$ have the areas ratio $\frac {[DBM]}{[BCM]} = \frac {BD^2}{BC^2} = \frac {AD}{AC}.$

Therefore $\frac {[DBM]}{[BCM]} = \frac {DE}{CE} = \frac {AD}{AC}. \blacksquare$

According the Cross-ratio criterion the four points $(A,C; E,D)$ are a harmonic range (on the real projective line).

vladimir.shelomovskii@gmail.com, vvsss

Symmetry and incircle

Symmetry incercle 2024 17.png

Let $\triangle ABC$ with incircle $\omega = \odot DEF, D \in BC, E \in AC, F\in AB$ be given.

Point $G = BC \cap EF.$

1. Let $P$ be the point in $\overset{\Large\frown} {DE}.$

Denote $Q = GP \cap \omega, K = \odot BQF \cap GF, L = \odot CPE \cap FE.$

Prove that $\frac {KE} {FL} = \frac {AB}{AC}.$

2. Let $K$ be the point in the segment $GF.$

Let $L$ be the point in the ray $FE,$ such that $\frac {KE} {FL} = \frac {AB}{AC}.$

Denote $Q = \odot BFK \cap \omega, P = \odot CEL \cap \omega.$

Prove that points $G, Q,$ and $P$ are collinear.

Proof

\[1. \angle GFQ = \angle GPE \implies \triangle GFQ \sim \triangle GPE \implies \frac {GF}{GP} = \frac {FQ}{PE}.\] \[\angle GPF = \angle GEQ \implies \triangle GFP \sim \triangle GQE \implies \frac {FP}{QE} = \frac {GP}{GE} \implies \frac {GF}{GE} = \frac {FQ \cdot FP}{PE \cdot QE}.\] We use Menelaus theorem for a triangle $\triangle AEF$ and a transversal line $GBC$ and get: \[\frac {AB \cdot GF \cdot CE}{BF \cdot GE \cdot AC} = 1 \implies \frac {AB}{AC}= \frac{GE \cdot BF}{CE \cdot GF}.\] \[\angle FKQ = \angle FBQ, \angle FEQ = \angle BFQ \implies \triangle KQE \sim \triangle BQF \implies \frac {KE}{BF} = \frac {QE}{QF}.\] \[\angle PFE = \angle PEC, \angle ELP = \angle ECP \implies \triangle LPF \sim \triangle CPE \implies \frac {LF}{CE} = \frac {FP}{EP}.\] \[\frac {KE}{LF} = \frac {BF \cdot QE \cdot EP}{QF \cdot CE \cdot FP} = \frac {BF \cdot GE}{CE \cdot GF} = \frac {AB}{AC}.\]

2. Denote $Q' = GP \cap \omega, K' = \odot BQ'F \cap GF.$ Then \[\frac {K'E}{LF} = \frac {AB}{AC} \implies K' = K \implies Q = Q' \in GP.\] vladimir.shelomovskii@gmail.com, vvsss

Symmetry and incircle A

Symmetry circle 2024 17.png

Denote $\omega = \odot DEF, P \in \omega$ is the arbitrary point. $Q = GP \cap \omega, Q \ne P.$

Prove that $BQ, CP,$ and $AD$ are concurrent.

Proof

Denote $I$ is the incenter of $\triangle ABC, I' = GI \cap AD.$

Let us make the projective transformations mapping circle $\omega$ onto circle and point $I'$ onto center of this circle.

Denote $X_0$ the result of transformation $T$ of point $X: T(X) = X_0.$ This transformation maps point $G$ to infinity.

Segment $AD$ this transformation maps onto diameter $\omega_0,$ onto $A_0D_0.$

We use the cross-ratio $\frac {BC \cdot GD}{BD \cdot GC}$ which is fixed, equation $\frac{T(GD)}{T(GC)} = 1,$ the Claim, and get

$\frac {B_0C_0}{B_0D_0} = 2,$ so $D_0$ is the midpoint of $B_0C_0 \implies A_0B_0 = A_0C_0.$

Point $G_0$ in infinity, so $B_0C_0 || Q_0P_0 || F_0E_0.$

Lines $B_0Q_0$ and $C_0P_0$ are crossing at the line of symmetry $A_0D_0,$ therefore lines $BQ, CP,$ and $AD$ are concurrent.

Claim

2024 17 B.png

Let $\triangle ABC$ with incircle $\omega = \odot DEF, D \in BC, E \in AC, F\in AB$ be given.

Point $G = BC \cap EF.$ Prove $\frac {BC \cdot GD} {BD \cdot GC}= 2.$

Proof

WLOG, $AB < AC.$ Denote $B'$ the point in $AC$ such that $AB' = AB.$ \[EB' = FB, BB' || EF \implies \frac {B'C }{BC} = \frac {B'E}{BG} \implies\] \[BG = \frac {BC \cdot B'E}{B'C} = \frac {BC \cdot BD}{AC - AB}.\] Denote $a= BC, b = AC, c = AB \implies BD = \frac {a+c - b}{2}, CD = \frac {a+b - c}{2},$ \[BG = \frac {a(a+c - b)}{2(b - c)}, CG = BG + BC = \frac {a(a + b - c)}{2(b - c)}\implies \frac {BC \cdot GD} {BD \cdot CG}= 2.\] vladimir.shelomovskii@gmail.com, vvsss

Symmetry for 60 degrees angle

60 symm.png
60 symm1.png

Let an isosceles triangle $\triangle ABC (\angle B = \angle C)$ be given.

Let $BE (E \in AC)$ be the bisector of $\angle B.$

a) $\angle B = \angle C = 40^\circ.$ Prove that $BE + AE = BC.$

b) $BE + AE = BC.$ Prove that $\angle B = 40^\circ.$

Proof

a) One can find successively angles (see diagram).

b) Let $\angle B = 4 \alpha \implies \angle C = 4 \alpha, \angle ABE = \angle CBE = 2 \alpha.$

Let $BE = BD, D \in BC \implies BC - BE = DC =AE.$

Let $EF||BC \implies \angle FEB = 2 \alpha = \angle FBE \implies EC = BF = FE.$

\[DC = AE, CE = FE, \angle AEF = \angle DCE \implies \triangle AEF = \triangle DCE \implies\] \[ED = CD \implies \angle CED = 4 \alpha.\] \[\angle BDE = 90^\circ - \alpha = \angle CED + \angle DCE = 8 \alpha \implies \alpha = 10 ^\circ \implies \angle B = 40^\circ.\]

vladimir.shelomovskii@gmail.com, vvsss

See also

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