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Isogonal conjugate: Difference between revisions

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== Definition of isogonal conjugate of a point ==
== Definition of isogonal conjugate of a point ==
[[File:Definitin 1.png|390px|right]]
[[File:Definitin 1.png|350px|right]]
Let triangle <math>\triangle ABC</math> be given. Let <math>\omega</math> be the circumcircle of <math>ABC.</math> Let point <math>P</math> be in the plane of <math>\triangle ABC, P \notin AB, P \notin BC, P \notin AC, P \notin \omega.</math>
Let triangle <math>\triangle ABC</math> be given. Let <math>\omega</math> be the circumcircle of <math>ABC.</math> Let point <math>P</math> be in the plane of <math>\triangle ABC, P \notin AB, P \notin BC, P \notin AC, P \notin \omega.</math>
Denote by <math>a,b,c</math> the lines <math>BC, CA, AB,</math> respectively. Denote by <math>p_a, p_b, p_c</math> the lines <math>PA</math>, <math>PB</math>, <math>PC</math>, respectively.
Denote by <math>a,b,c</math> the lines <math>BC, CA, AB,</math> respectively. Denote by <math>p_a, p_b, p_c</math> the lines <math>PA</math>, <math>PB</math>, <math>PC</math>, respectively.
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==Equidistant isogonal conjugate points==
==Equidistant isogonal conjugate points==
[[File:Equal distances.png|330px|right]]
[[File:Equal distances.png|300px|right]]
[[File:Equidistant points.png|330px|right]]
[[File:Equidistant points.png|300px|right]]
Let triangle <math>ABC</math> with incenter <math>I</math> be given.
Let triangle <math>ABC</math> with incenter <math>I</math> be given.
Denote <math>\omega = \odot BIC.</math>
Denote <math>\omega = \odot BIC.</math>
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Point <math>P'</math> is the isogonal conjugate of the point <math>P</math> with respect to <math>\triangle ABC \implies</math>
Point <math>P'</math> is the isogonal conjugate of the point <math>P</math> with respect to <math>\triangle ABC \implies</math>
<cmath>\angle PBI = \angle P'BI, \angle PCI = \angle P'CI \implies \angle P'BI = \angle P'CI.</cmath>
<cmath>\angle PBI = \angle P'BI, \angle PCI = \angle P'CI \implies</cmath>
<cmath>\angle P'BI = \angle P'CI.</cmath>
So points <math>B,C,I, P,</math> and <math>P'</math> are concyclic.
So points <math>B,C,I, P,</math> and <math>P'</math> are concyclic.


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==Simplified distance formula for isogonal points==
==Simplified distance formula for isogonal points==
[[File:1 pare and.png|370px|right]]
[[File:1 pare and.png|330px|right]]
Let triangle <math>\triangle ABC,</math> points <math>P</math> and <math>P',</math> and <math>\odot ABC = \Omega</math> be given. Let point <math>P'</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math>
Let triangle <math>\triangle ABC,</math> points <math>P</math> and <math>P',</math> and <math>\odot ABC = \Omega</math> be given. Let point <math>P'</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math>
<cmath>D = AP \cap BC, E = AP' \cap BC, F = AP \cap \Omega, G = AP' \cap \Omega.</cmath>  
<cmath>D = AP \cap BC, E = AP' \cap BC,</cmath>
<cmath>F = AP \cap \Omega, G = AP' \cap \Omega.</cmath>  
Prove that <math>PF \cdot P'G= AF \cdot EG.</math>
Prove that <math>PF \cdot P'G= AF \cdot EG.</math>


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==Point on circumcircle==
==Point on circumcircle==
[[File:RADAX.png|350px|right]]
[[File:RADAX.png|330px|right]]
Let triangle <math>\triangle ABC,</math> points <math>D \in BC</math> and <math>E \in BC</math> be given.
Let triangle <math>\triangle ABC,</math> points <math>D \in BC</math> and <math>E \in BC</math> be given.


Denote <math>\Omega = \odot ABC, \omega = \odot AED, G = \omega \cap \Omega \ne A, F = AG \cap BC,</math>
Denote <math>\Omega = \odot ABC, \omega = \odot AED,</math>
<cmath>G = \omega \cap \Omega \ne A, F = AG \cap BC,</cmath>
<cmath>K = AE \cap \Omega \ne A, L = GD \cap \Omega \ne G.</cmath>
<cmath>K = AE \cap \Omega \ne A, L = GD \cap \Omega \ne G.</cmath>
Prove that <math>KL || BC.</math>  
Prove that <math>KL || BC.</math>  
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<math>AL</math> is the isogonal conjugate to <math>AK</math> with respect <math>\angle BAC.</math>
<math>AL</math> is the isogonal conjugate to <math>AK</math> with respect <math>\angle BAC.</math>


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==Fixed point on circumcircle==
==Fixed point on circumcircle==
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Spiral similarity centered at <math>A</math> which maps <math>\Omega</math> into <math>\odot AYD</math> transform point <math>H</math> into point <math>D \implies</math>
Spiral similarity centered at <math>A</math> which maps <math>\Omega</math> into <math>\odot AYD</math> transform point <math>H</math> into point <math>D \implies</math>
<cmath>\overset{\Large\frown} {ACH} =  \overset{\Large\frown} {AD} \implies \angle AGH = \angle AYD \implies GH||BC \implies</cmath> <cmath>\overset{\Large\frown} {CH} =  \overset{\Large\frown} {BG} \implies \angle BAG = \angle CAH \implies H \in AP'.</cmath>
<cmath>\overset{\Large\frown} {ACH} =  \overset{\Large\frown} {AD} \implies</cmath>
<cmath>\angle AGH = \angle AYD \implies GH||BC \implies</cmath>  
<cmath>\overset{\Large\frown} {CH} =  \overset{\Large\frown} {BG} \implies</cmath>
<cmath>\angle BAG = \angle CAH \implies H \in AP'.</cmath>
Points <math>F', H,</math> and <math>D</math> are collinear.
Points <math>F', H,</math> and <math>D</math> are collinear.


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==Miquel point for isogonal conjugate points==
==Miquel point for isogonal conjugate points==
[[File:Miquel of one pare.png|370px|right]]
[[File:Miquel of one pare.png|330px|right]]
[[File:Miquel 1 pare.png|370px|right]]
[[File:Miquel 1 pare.png|330px|right]]
Let triangle <math>\triangle ABC,</math> points <math>Q \in BC</math> and <math>P</math> be given. Let point <math>P'</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math>  
Let triangle <math>\triangle ABC,</math> points <math>Q \in BC</math> and <math>P</math> be given. Let point <math>P'</math> be the isogonal conjugate of a point <math>P</math> with respect to a triangle <math>\triangle ABC.</math>  
<cmath>D = QP \cap AP', E = AP \cap QP'.</cmath>
<cmath>D = QP \cap AP', E = AP \cap QP'.</cmath>
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Points <math>P</math> and <math>E</math> lies on the same line, therefore <cmath>PF \cdot P'G = EF \cdot DG \implies</cmath>
Points <math>P</math> and <math>E</math> lies on the same line, therefore <cmath>PF \cdot P'G = EF \cdot DG \implies</cmath>


<cmath>\frac {PF-EP}{PF} = \frac{DG-DP'}{DG} \implies \frac {EP}{PF} = \frac{DP'}{DG}.</cmath>
<cmath>\frac {PF-EP}{PF} = \frac{DG-DP'}{DG} \implies</cmath>
<cmath>\frac {EP}{PF} = \frac{DP'}{DG}.</cmath>
Point <math>M</math> lies on circles <math>APD</math> and <math>AEP' \implies </math> spiral similarity centered at <math>M</math> transform triangle <math>\triangle MPE</math> to <math>\triangle MDP' \implies</math>
Point <math>M</math> lies on circles <math>APD</math> and <math>AEP' \implies </math> spiral similarity centered at <math>M</math> transform triangle <math>\triangle MPE</math> to <math>\triangle MDP' \implies</math>
<cmath>\angle MPE = \angle MDP', \frac {MP}{MD} = \frac {PE}{DP'} = \frac{PF}{DG} \implies</cmath>
<cmath>\angle MPE = \angle MDP', \frac {MP}{MD} = \frac {PE}{DP'} = \frac{PF}{DG} \implies</cmath>
<cmath>\triangle MPF \sim \triangle MDG \implies</cmath> <cmath>\angle AFM = \angle AGM \implies M \in \odot ABC.</cmath>
<cmath>\triangle MPF \sim \triangle MDG \implies</cmath> <cmath>\angle AFM = \angle AGM \implies M \in \odot ABC.</cmath>


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==Point on circumcircle==
==Point on circumcircle==
[[File:Point on circumcircle 0.png|400px|right]]
[[File:Point on circumcircle 0.png|400px|right]]
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Prove that <math>X = EF \cap HG \in \Omega.</math>
Prove that <math>X = EF \cap HG \in \Omega.</math>
[[File:Point on circumcircle 1.png|400px|right]]
[[File:Point on circumcircle 1.png|320px|right]]


<i><b>Proof, Simplified problem</b></i>
<i><b>Proof, Simplified problem</b></i>
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<i><b>Proof</b></i>
<i><b>Proof</b></i>
[[File:Point on circumcircle 2.png|400px|right]]
[[File:Point on circumcircle 2.png|350px|right]]
Let points <math>P'</math> and <math>Q'</math> be the isogonal conjugate of a points <math>P</math> and <math>Q</math> with respect to a triangle <math>\triangle ABC, \omega' = \odot Q'PD, \theta' = \odot P'QD.</math>
Let points <math>P'</math> and <math>Q'</math> be the isogonal conjugate of a points <math>P</math> and <math>Q</math> with respect to a triangle <math>\triangle ABC, \omega' = \odot Q'PD, \theta' = \odot P'QD.</math>


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points <math>M, H, E, X</math> are concyclic on <math>\Omega.</math>
points <math>M, H, E, X</math> are concyclic on <math>\Omega.</math>


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==Isogonal of line BC with respect to angle BAC==
==Isogonal of line BC with respect to angle BAC==
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==Isogonal bijection lines and points==
==Isogonal bijection lines and points==
[[File:Isogonal of l.png|350px|right]]
[[File:Isogonal of l.png|330px|right]]
Let triangle <math>\triangle ABC</math> and line <math>\ell, P \in \ell</math> be given, <math>\Omega = \odot ABC.</math>
Let triangle <math>\triangle ABC</math> and line <math>\ell, P \in \ell</math> be given, <math>\Omega = \odot ABC.</math>


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<i><b>Proof</b></i>
<i><b>Proof</b></i>


WLOG, the configuration is the same as shown on diagram, <math>F = \ell \cap BC, AD' || \ell,  \theta = \angle PFB, AD || BC, AE</math> is the tangent to <math>\Omega.</math>
WLOG, the configuration is the same as shown on diagram, <math>F = \ell \cap BC, AD' || \ell,  \theta = \angle PFB, AD || BC,</math>
<math>AE</math> is the tangent to <math>\Omega.</math>


<math>AD</math> is isogonal to <math>AE, AD'</math> is isogonal to <math>AG</math> with respect to <math>\angle BAC \implies</math>
<math>AD</math> is isogonal to <math>AE, AD'</math> is isogonal to <math>AG</math> with respect to <math>\angle BAC \implies</math>
Line 745: Line 752:
A bijection has been established between the set of lines parallel to a given one and the set of points of the circumcircle.
A bijection has been established between the set of lines parallel to a given one and the set of points of the circumcircle.


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==Miquel point for two pare isogonal points==
==Miquel point for two pare isogonal points==
[[File:2 pare Miquel o.png|400px|right]]
[[File:2 pare Miquel o.png|340px|right]]
Let triangle <math>\triangle ABC</math> and points <math>P</math> and <math>Q</math> be given.
Let triangle <math>\triangle ABC</math> and points <math>P</math> and <math>Q</math> be given.


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<cmath>\angle EMF = \angle RME - \angle RMF = \angle RQ'E - \angle RP'F = \angle P'Q'E - \angle DP'Q' = \angle P'DQ' = \angle EDF \implies M \in \odot DEF  \blacksquare</cmath>
<cmath>\angle EMF = \angle RME - \angle RMF = \angle RQ'E - \angle RP'F = \angle P'Q'E - \angle DP'Q' = \angle P'DQ' = \angle EDF \implies M \in \odot DEF  \blacksquare</cmath>


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==Isogonic center’s conjugate point==
==Isogonic center’s conjugate point==
[[File:Fermat 1.png|500px|right]]
[[File:Fermat 1.png|400px|right]]
[[File:Fermat 2.png|300px|right]]
[[File:Fermat 2.png|300px|right]]
Let triangle <math>ABC</math> with isogonic center <math>F (X(13)</math> or <math>X(14))</math> be given. Denote <math>\omega = \odot BIC.</math>
Let triangle <math>ABC</math> with isogonic center <math>F (X(13)</math> or <math>X(14))</math> be given. Denote <math>\omega = \odot BIC.</math>
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Let <math>I</math> be the incenter of <math>\triangle ABC, F =X(13).</math>
Let <math>I</math> be the incenter of <math>\triangle ABC, F =X(13).</math>
<cmath>A_1 = AF \cup BC, E = AI \cup BC, D \in BC, AD \perp AE, \omega = \odot AED.</cmath>
<cmath>A_1 = AF \cup BC, E = AI \cup BC, D \in BC,</cmath>
<cmath>AD \perp AE, \omega = \odot AED.</cmath>
Let <math>F_1 = AF \cup \omega, F'</math> is simmetric to <math>F_1</math> with respect <math>BC \implies A_1F' = \ell_A.</math>
Let <math>F_1 = AF \cup \omega, F'</math> is simmetric to <math>F_1</math> with respect <math>BC \implies A_1F' = \ell_A.</math>


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The second diagram show construction in the case <math>F =X(14).</math> The proof is similar.
The second diagram show construction in the case <math>F =X(14).</math> The proof is similar.


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==Three pairs isogonal points==
==Three pairs isogonal points==
[[File:Shar 2024 20.png|400px|right]]
[[File:Shar 2024 20.png|350px|right]]
Let a triangle <math>ABC,</math> points <math>D</math> and <math>E \in AD</math> be given, <math>F = CD \cap BE.</math>
Let a triangle <math>ABC,</math> points <math>D</math> and <math>E \in AD</math> be given, <math>F = CD \cap BE.</math>
Points <math>D', E'</math> and <math>F'</math> are the isogonal conjugate of the points <math>D, E,</math> and <math>F,</math> respectively, with respect to <math>\triangle ABC.</math>  
Points <math>D', E'</math> and <math>F'</math> are the isogonal conjugate of the points <math>D, E,</math> and <math>F,</math> respectively, with respect to <math>\triangle ABC.</math>  
Line 817: Line 824:
We multiply these equations and get
We multiply these equations and get
<cmath>\frac {AE \cdot BF \cdot CD}{AD \cdot BE \cdot CF} = \frac{AD' \cdot BE' \cdot CF'}{AE' \cdot BF' \cdot CD'} = \frac {\sin \varphi_A \cdot \sin \varphi_B \cdot \sin \varphi_C}{\sin (\alpha - \varphi_A) \cdot \sin (\beta - \varphi_B) \cdot \sin (\gamma - \varphi_C)}.</cmath>
<cmath>\frac {AE \cdot BF \cdot CD}{AD \cdot BE \cdot CF} = \frac{AD' \cdot BE' \cdot CF'}{AE' \cdot BF' \cdot CD'} = \frac {\sin \varphi_A \cdot \sin \varphi_B \cdot \sin \varphi_C}{\sin (\alpha - \varphi_A) \cdot \sin (\beta - \varphi_B) \cdot \sin (\gamma - \varphi_C)}.</cmath>
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==Ratio for three pairs of isogonal points==
==Ratio for three pairs of isogonal points==
[[File:Shar 2024 20 1.png|400px|right]]
[[File:Shar 2024 20 1.png|350px|right]]
[[File:Shar 2024 20 2.png|400px|right]]
[[File:Shar 2024 20 2.png|350px|right]]
Let a triangle <math>ABC,</math> points <math>D</math> and <math>E \in AD</math> be given, <math>F = CD \cap BE.</math>
Let a triangle <math>ABC,</math> points <math>D</math> and <math>E \in AD</math> be given, <math>F = CD \cap BE.</math>


Line 834: Line 842:


<math>[DEF] = 1,</math> where <math>[X]</math> is the area of the figure <math>X.</math>
<math>[DEF] = 1,</math> where <math>[X]</math> is the area of the figure <math>X.</math>
<cmath>\frac {[ADF]}{[DEF]} = \frac {AD}{DE} = u, \frac {[ACF]}{[DEF]} = \frac {[ACF]}{[ADF]} \cdot \frac {[ADF]}{[DEF]} = uv.</cmath>
<cmath>\frac {[ADF]}{[DEF]} = \frac {AD}{DE} = u,</cmath>
Similarly, <cmath>\frac {[CFE]}{[DEF]} = v, \frac {[BCE]}{[DEF]} = vw, \frac {[BDE]}{[DEF]} = w, \frac {[ABD]}{[DEF]} = uw.</cmath>
<cmath>\frac {[ACF]}{[DEF]} = \frac {[ACF]}{[ADF]} \cdot \frac {[ADF]}{[DEF]} = uv.</cmath>
Similarly, <math>\frac {[CFE]}{[DEF]} = v, \frac {[BCE]}{[DEF]} = vw,</math>
<cmath>\frac {[BDE]}{[DEF]} = w, \frac {[ABD]}{[DEF]} = uw.</cmath>
<cmath>\frac {[ABC]}{[DEF]} = 1 + u + v + w + uv + uw + vw = (1 + u)(1 + v)(1 + w) - uvw,</cmath>
<cmath>\frac {[ABC]}{[DEF]} = 1 + u + v + w + uv + uw + vw = (1 + u)(1 + v)(1 + w) - uvw,</cmath>
<cmath>\frac {[ABC]}{[DEF]}  = \frac {AE \cdot BF \cdot CD}{DE \cdot DF \cdot EF} -  \frac {AD \cdot BE \cdot CF}{DE \cdot DF \cdot EF}.</cmath>
<cmath>\frac {[ABC]}{[DEF]}  = \frac {AE \cdot BF \cdot CD}{DE \cdot DF \cdot EF} -  \frac {AD \cdot BE \cdot CF}{DE \cdot DF \cdot EF}.</cmath>
Line 845: Line 855:
Comment: The main idea of the proof was found by Leonid Shatunov.
Comment: The main idea of the proof was found by Leonid Shatunov.


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== Problems ==
== Problems ==

Latest revision as of 16:42, 20 October 2025

The isogonal theorem

Isogonal lines definition

Let a line $\ell$ and a point $O$ lying on $\ell$ be given. A pair of lines symmetric with respect to $\ell$ and containing the point $O$ be called isogonals with respect to the pair $(\ell,O).$

Sometimes it is convenient to take one pair of isogonals as the base one, for example, $OA$ and $OB$ are the base pair. Then we call the remaining pairs as isogonals with respect to the angle $\angle AOB.$

Projective transformation

It is known that the transformation that maps a point with coordinates $(x,y)$ into a point with coordinates $(\frac{1}{x}, \frac {y}{x}),$ is projective.

If the abscissa axis coincides with the line $\ell$ and the origin coincides with the point $O,$ then the isogonals define the equations $y = \pm kx,$ and the lines $(\frac{1}{x}, \pm k)$ symmetrical with respect to the line $\ell$ become their images.

It is clear that, under the converse transformation (also projective), such pairs of lines become isogonals, and the points equidistant from $\ell$ lie on the isogonals.

The isogonal theorem

Let two pairs of isogonals $OX - OX'$ and $OY - OY'$ with respect to the pair $(\ell,O)$ be given. Denote $Z = XY \cap X'Y', Z' = X'Y \cap XY'.$

Prove that $OZ$ and $OZ'$ are the isogonals with respect to the pair $(\ell,O).$

Proof

Let us perform a projective transformation of the plane that maps the point $O$ into a point at infinity and the line $\ell$ maps to itself. In this case, the isogonals turn into a pair of straight lines parallel to $\ell$ and equidistant from $\ell.$

The converse (also projective) transformation maps the points equidistant from $\ell$ onto isogonals. We denote the image and the preimage with the same symbols.

Let the images of isogonals are vertical lines. Let coordinates of images of points be \[X(-a, 0), X'(a,u), Y(-b,v), Y'(b,w).\] Equation of a straight line $XY$ is $\frac{x + a}{a - b} = \frac {y}{v}.$

Equation of a straight line $X'Y'$ is $\frac{x - a}{b - a} = \frac {y - u}{w - u}.$

The abscissa $Z_x$ of the point $Z$ is $Z_x = \frac {v a - a w + u b}{u - v - w}.$

Equation of a straight line $XY'$ is $\frac{x + a}{b + a} = \frac {y}{w}.$

Equation of a straight line $X'Y$ is $\frac{x - a}{- b - a} = \frac {y - u}{v - u}.$

The abscissa $Z'_x$ of the point $Z'$ is $Z'_x = \frac {v a - a w + u b}{- u + v + w} = - Z_x \implies$

Preimages of the points $Z$ and $Z'$ lie on the isogonals. $\blacksquare$

The isogonal theorem in the case of parallel lines

Let $OY$ and $OY'$ are isogonals with respect $\angle XOX'.$

Let lines $XY$ and $X'Y'$ intersect at point $Z, X'Y || XY'.$

Prove that $OZ$ and line $l$ through $O$ parallel to $XY'$ are the isogonals with respect $\angle XOX'.$

Proof

The preimage of $Z'$ is located at infinity on the line $l.$

The equality $Z'_x = -Z_x$ implies the equality the slopes modulo of $OZ$ and $l$ to the bisector of $\angle XOX'. \blacksquare$

Converse theorem

Let lines $XY$ and $X'Y'$ intersect at point $Z, X'Y || XY'.$

Let $OZ$ and $l$ be the isogonals with respect $\angle XOX'.$

Prove that $OY$ and $OY'$ are isogonals with respect $\angle XOX' (\angle XOY' = \angle YOX').$

Proof

The preimage of $Z'$ is located at infinity on the line $l,$ so the slope of $OZ$ is known.

Suppose that $y' \in XY', y' \ne Y', \angle y'OX = \angle YOX'.$

The segment $XY$ and the lines $XY', OZ$ are fixed $\implies$

$y'X'$ intersects $XY$ at $z \ne Z,$

but there is the only point where line $OZ$ intersect $XY.$ Сontradiction. $\blacksquare$

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Parallel segments

Let triangle $ABC$ be given. Let $AD$ and $AE$ be the isogonals with respect $\angle BAC.$ Let $BD ||CE, P = BE \cap CD.$

Prove that $P$ lies on bisector of $\angle BAC$ and $BD||AP.$

Proof

Both assertions follow from The isogonal theorem in the case of parallel lines

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Perpendicularity

Let triangle $ABC$ be given. Right triangles $ABD$ and $ACE$ with hypotenuses $AD$ and $AE$ are constructed on sides $AB$ and $AC$ to the outer (inner) side of $\triangle ABC.$ Let $\angle BAD = \angle CAE, H = CD \cap BE.$ Prove that $AH \perp BC.$

Proof

Let $\ell$ be the bisector of $\angle BAC, F = BD \cap CE.$

$AB$ and $AC$ are isogonals with respect to the pair $(\ell,A).$

$AD$ and $AE$ are isogonals with respect to the pair $(\ell,A) \implies$

$AH$ and $AF$ are isogonals with respect to the pair $(\ell,A)$ in accordance with The isogonal theorem.

$\angle ABD = \angle ACE = 90^\circ \implies$

$AF$ is the diameter of circumcircle of $\triangle ABC.$

Circumradius and altitude are isogonals with respect bisector and vertex of triangle, so $AH \perp BC.$ $\blacksquare$

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Fixed point

Let fixed triangle $ABC$ be given. Let points $D$ and $E$ on sidelines $BC$ and $AB,$ respectively be the arbitrary points.

Let $F$ be the point on sideline $AC$ such that $\angle BDE = \angle CDF.$

$G = BF \cap CE.$ Prove that line $DG$ pass through the fixed point.

Proof

We will prove that point $A',$ symmetric $A$ with respect $\ell = BC,$ lies on $DG$.

$\angle BDE = \angle CDF \implies DE$ and $DF$ are isogonals with respect to $(\ell, D).$

$A = BE \cap CF \implies$ points $A$ and $G$ lie on isogonals with respect to $(\ell, D)$ in accordance with The isogonal theorem.

Point $A'$ symmetric $A$ with respect $\ell$ lies on isogonal $AD$ with respect to $(\ell, D),$ that is $DG.$ $\blacksquare$

vvsss

Bisector

Let a convex quadrilateral $ABCD$ be given. Let $I$ and $J$ be the incenters of triangles $\triangle ABC$ and $\triangle ADC,$ respectively.

Let $I'$ and $J'$ be the A-excenters of triangles $\triangle ABC$ and $\triangle ADC,$ respectively. $E = IJ' \cap I'J.$

Prove that $CE$ is the bisector of $\angle BCD.$

Proof

$\angle ICI' = \angle JCJ' = 90^\circ \implies$

$CI'$ and $CJ'$ are isogonals with respect to the angle $\angle ICJ.$

$A = II' \cap JJ' \implies AC$ and $EC$ are isogonals with respect to the angle $\angle ICJ$ in accordance with The isogonal theorem.

Denote $\angle ACI = \angle BCI = \alpha, \angle ACJ = \angle DCJ = \beta.$

WLOG, $\beta \ge \alpha.$ \[\angle ACJ = \angle ACE + \alpha = \beta,\] \[\angle BCE = 2 \alpha + \beta - \alpha = \alpha + \beta = \angle DCE. \blacksquare\]

vvsss

Isogonal of the diagonal of a quadrilateral

Given a quadrilateral $ABCD$ and a point $P$ on its diagonal such that $\angle APB = \angle APD.$

Let $E = AB \cap CD, F = AD \cap BC.$

Prove that $\angle BPE = \angle DPF.$

Proof

Let us perform a projective transformation of the plane that maps the point $P$ to a point at infinity and the line $\ell = AC$ into itself.

In this case, the images of points $B$ and $D$ are equidistant from the image of $AC \implies$

the point $M$ (midpoint of $BD)$ lies on $\ell \implies$

$AC$ contains the midpoints of $AC$ and $BD \implies$

$\ell$ is the Gauss line of the complete quadrilateral $ABCDEF \implies$ $\ell$ bisects $EF \implies EE_0 = FF_0 \implies$

the preimages of the points $E$ and $F$ lie on the isogonals $PE$ and $PF. \blacksquare$

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Isogonals in trapezium

Let the trapezoid $AEFC, AC||EF,$ be given. Denote \[B = AE \cap CF, D = AF \cap CE.\]

The point $M$ on the smaller base $AC$ is such that $EM = MF.$

Prove that $\angle AMB = \angle AMD.$

Proof

\[EM = MF \implies\] \[\angle AME = \angle MEF = \angle MFE = \angle CMF.\] Therefore $EM$ and $FM$ are isogonals with respect $(AC,M).$

Let us perform a projective transformation of the plane that maps the point $M$ to a point at infinity and the line $\ell = AC$ into itself.

In this case, the images of points $E$ and $F$ are equidistant from the image of $\ell \implies AC$ contains the midpoints of $AC$ and $EF$, that is, $\ell$ is the Gauss line of the complete quadrilateral $ABCDEF \implies$

$\ell$ bisects $BD \implies BB_0 = DD_0 \implies$

The preimages of the points $B$ and $D$ lie on the isogonals $MB$ and $MD. \blacksquare$

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Isogonals in complete quadrilateral

Let complete quadrilateral $ABCDEF (E = AB \cap CD, F = AC \cap BD)$ be given. Let $M$ be the Miquel point of $ABCD.$

Prove that $AM$ is isogonal to $DM$ and $EM$ is isogonal to $FM$ with respect $\angle BMC.$

Proof \[\angle BME = \angle BDE = \angle CDF = \angle CMF.\] \[\angle BMD = \angle BED = \angle AEC = \angle AMC. \blacksquare\]

Isogonal of the bisector of the triangle

The triangle $ABC$ be given. The point $D$ chosen on the bisector $AA'.$

Denote \[B' = BD \cap AC, C' = CD \cap AB,\] \[E = BB' \cap A'C', F = CC' \cap A'B'.\] Prove that $\angle BAE = \angle CAF.$

Proof

Let us perform a projective transformation of the plane that maps the point $A$ to a point at infinity and the line $\ell = AA'$ into itself.

In this case, the images of segments $BC'$ and $B'C$ are equidistant from the image of $\ell \implies BC' || B'C || \ell.$

Image of point $D$ is midpoint of image $BB'$ and midpoint image $CC' \implies$

Image $BCB'C'$ is parallelogramm $\implies$

$BC = B'C' \implies \frac {DE}{BD} = \frac {DF}{CD} \implies$ distances from $E$ and $F$ to $\ell$ are equal $\implies$

Preimages $AE$ and $AF$ are isogonals with respect $(\ell,A). \blacksquare$

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Points on isogonals

The triangle $ABC$ be given. The point $D$ chosen on $BC.$ The point $E$ chosen on $BC$ such that $AD$ and $AE$ are isogonals with respect $\angle BAC.$

Prove that $\frac {AB^2}{BD \cdot BE} = \frac{ AC^2}{CD \cdot CE}.$

Proof

Denote $\angle BAD = \angle CAE = \varphi,$ $\angle B = \beta, \angle C = \gamma \implies$ $\angle ADE = \beta + \varphi, \angle AED = \gamma + \varphi, \angle BAE = \angle CAD = \psi+\varphi.$

We use the Law of Sines and get: \[\frac {AB}{BD} = \frac {\sin (\beta + \varphi)}{\sin \varphi}, \frac {AB}{BE} = \frac {\sin (\gamma + \varphi)}{\sin (\psi +\varphi)},\] \[\frac {AC}{CE} = \frac {\sin (\gamma + \varphi)}{\sin \varphi}, \frac {AC}{CD} = \frac {\sin (\beta + \varphi)}{\sin (\psi +\varphi)} \implies\] \[\frac {AB \cdot AB}{BD \cdot BE} = \frac {AC \cdot AC}{CD \cdot CE} = \frac {\sin (\beta + \varphi) \cdot \sin (\gamma +\varphi)} {\sin \varphi \cdot \sin (\psi +\varphi)}.\]

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Trapezoid

The lateral side $CD$ of the trapezoid $ABCD$ is perpendicular to the bases, point $P$ is the intersection point of the diagonals $ABCD$.

Point $Q$ is taken on the circumcircle $\omega$ of triangle $PCD$ diametrically opposite to point $P.$

Prove that $\angle BQC = \angle AQD.$

Proof

WLOG, $CD$ is not the diameter of $\omega.$ Let sidelines $AD$ and $BC$ intersect $\omega$ at points $D'$ and $C',$ respectively.

$DD' \perp CD, CC' \perp CD \implies CDD'C'$ is rectangle $\implies$ $CC' = DD' \implies \angle CQC' = \angle DQD'.$

$QE||BC$ is isogonal to $QO$ with respect $\angle CQD \implies$

$QE||BC$ is isogonal to $QP$ with respect $\angle CQD \implies$

In accordance with The isogonal theorem in case parallel lines $\angle DQO = \angle CQE.$

$QE||BC$ is isogonal to $QP$ with respect $\angle CQD, P = AC \cap BD \implies$

$\angle AQD = \angle BQC$ in accordance with Converse theorem for The isogonal theorem in case parallel lines. $\blacksquare$

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Definition of isogonal conjugate of a point

Let triangle $\triangle ABC$ be given. Let $\omega$ be the circumcircle of $ABC.$ Let point $P$ be in the plane of $\triangle ABC, P \notin AB, P \notin BC, P \notin AC, P \notin \omega.$ Denote by $a,b,c$ the lines $BC, CA, AB,$ respectively. Denote by $p_a, p_b, p_c$ the lines $PA$, $PB$, $PC$, respectively. Denote by $q_a$, $q_b$, $q_c$ the reflections of $p_a$, $p_b$, $p_c$ over the angle bisectors of angles $A$, $B$, $C$, respectively.

Prove that lines $q_a$, $q_b$, $q_c$ concur at a point $Q.$ This point is called the isogonal conjugate of $P$ with respect to triangle $ABC$.

Proof

By our constructions of the lines $q$, $\angle p_a b \equiv \angle q_a c$, and this statement remains true after permuting $a,b,c$. Therefore by the trigonometric form of Ceva's Theorem \[\frac{\sin \angle q_a b}{\sin \angle c q_a} \cdot \frac{\sin \angle q_b c}{\sin \angle a q_b} \cdot \frac{\sin \angle q_c a}{\sin \angle b q_c} = \frac{\sin \angle p_a c}{\sin \angle b p_a} \cdot \frac{\sin \angle p_b a}{\sin \angle c p_b} \cdot \frac{\sin \angle p_c b}{\sin \angle a p_c} = 1,\] so again by the trigonometric form of Ceva, the lines $q_a, q_b, q_c$ concur, as was to be proven. $\blacksquare$

Corollary

Let points P and Q lie on the isogonals with respect angles $\angle B$ and $\angle C$ of triangle $\triangle ABC.$

Then these points lie on isogonals with respect angle $\angle A.$

Corollary 2

Let point $P$ be in the sideline $BC$ of $\triangle ABC, P \ne B, P \ne C.$

Then the isogonal conjugate of a point $P$ is a point $A.$

Points $A,B,$ and $C$ do not have an isogonally conjugate point.

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Three points

Let fixed triangle $ABC$ be given. Let the arbitrary point $D$ not be on sidelines of $\triangle ABC.$ Let $E$ be the point on isogonal of $CD$ with respect angle $\angle ACB.$ Let $F$ be the crosspoint of isogonal of $BD$ with respect angle $\angle ABC$ and isogonal of $AE$ with respect angle $\angle BAC.$

Prove that lines $AD, BE,$ and $CF$ are concurrent.

Proof

Denote $D' = BF \cap CE, S = CF \cap BE.$

$AE$ and $AF$ are isogonals with respect $\angle BAC \implies$

$D'$ and S lie on isogonals of $\angle BAC.$

$\angle DBC = \angle D'BA, \angle DCB = \angle D'CA \implies$

$D'$ is isogonal conjugated of $D$ with respect $\triangle ABC \implies$

$D'$ and $D$ lie on isogonals of $\angle BAC.$

Therefore points $A, S$ and $D$ lie on the same line which is isogonal to $AD'$ with respect $\angle BAC. \blacksquare$

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Second definition

Let triangle $\triangle ABC$ be given. Let point $P$ lies in the plane of $\triangle ABC,$ \[P \notin AB, P \notin BC, P \notin AC.\] Let the reflections of $P$ in the sidelines $BC, CA, AB$ be $P_1, P_2, P_3.$

Then the circumcenter $Q$ of the $\triangle P_1P_2P_3$ is the isogonal conjugate of $P.$

Points $A, B,$ and $C$ have not isogonal conjugate points.

Another points of sidelines $BC, AC, AB$ have points $A, B, C,$ respectively as isogonal conjugate points.

Proof \[PC = P_1C, PC = P_2C \implies P_1C = P_2C.\] \[\angle ACQ = \angle BCP_1 \implies \angle QCP_1 = \angle ACB.\] \[\angle BCQ = \angle ACP_2 \implies \angle QCP_2 = \angle ACB.\] $\angle QCP_1 = \angle QCP_2, CP_1 = CP_2, QC$ is common therefore \[\triangle QCP_1 = \triangle QCP_2 \implies QP_1 = QP_2.\] Similarly $QP_1 = QP_3 \implies Q$ is the circumcenter of the $\triangle P_1P_2P_3.$ $\blacksquare$

From definition 1 we get that $P$ is the isogonal conjugate of $Q.$

It is clear that each point $P$ has the unique isogonal conjugate point.

Let point $P$ be the point with barycentric coordinates $(p : q : r),$ \[p = [(P - B),(P - C)], q = [(P - C),(P - A)], r = [(P - A),(P - B)].\] Then $Q$ has barycentric coordinates \[(p' : q' : r'), p' = \frac {|B - C|^2}{p}, q' = \frac {|A-C|^2}{q}, r' = \frac {|A - B|^2}{r}.\]

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Distance to the sides of the triangle

Let $Q$ be the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC.$

Let $E$ and $D$ be the projection $P$ on sides $AC$ and $BC,$ respectively.

Let $E'$ and $D'$ be the projection $Q$ on sides $AC$ and $BC,$ respectively.

Then $\frac {PE}{PD} = \frac{QD'}{QE'}.$

Proof

Let $\theta = \angle ACP = \angle BCQ, \Theta = \angle ACQ = \angle BCP.$ \[\frac {PE}{PD} = \frac {PC \sin \theta}{PC \sin \Theta} = \frac {QC \sin \theta}{QC \sin \Theta} = \frac {QD'}{QE'}. \blacksquare\] vladimir.shelomovskii@gmail.com, vvsss

Sign of isogonally conjugate points

Let triangle $\triangle ABC$ and points $P$ and $Q$ inside it be given.

Let $D, E, F$ be the projections $P$ on sides $BC, AC, AB,$ respectively.

Let $D', E', F'$ be the projections $Q$ on sides $BC, AC, AB,$ respectively.

Let $\frac {PE}{PD} = \frac{QD'}{QE'}, \frac {PF}{PD} = \frac{QD'}{QF'}.$ Prove that point $Q$ is the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC.$

One can prove a similar theorem in the case $P$ outside $\triangle ABC.$

Proof

\[\frac {PE}{PD} = \frac {PE}{PC} : \frac {PD}{PC} = \frac {\sin \angle ACP}{\sin \angle BCP},\] \[\frac {QD'}{QE'} = \frac {QD'}{QC} : \frac {QE'}{QC} = \frac {\sin \angle BCQ}{\sin \angle ACQ}.\]

Denote $\angle ACP = \varphi, \angle BCQ = \psi, \angle ACB = \gamma.$ \[\sin \varphi \cdot \sin (\gamma - \psi) = \sin \psi \cdot \sin (\gamma - \varphi) \implies\] \[\cos (\varphi - \gamma + \psi) - \cos(\varphi + \gamma - \psi) = \cos (\psi - \gamma + \varphi) - \cos(\psi + \gamma - \varphi)\] \[\cos (\gamma + \varphi - \psi) = \cos(\gamma - \psi + \varphi) \implies\] \[\cos \gamma \cos (\varphi - \psi) - \sin \gamma \sin (\varphi - \psi) = \cos \gamma \cos (\varphi - \psi) + \sin \gamma \sin (\varphi - \psi)\] \[2 \sin \gamma \cdot \sin (\varphi - \psi) = 0, \varphi + \psi < 180^\circ \implies \varphi = \psi.\] Similarly $\angle ABP = \angle CBQ.$ Hence point $Q$ is the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC. \blacksquare$

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Circumcircle of pedal triangles

Let $Q$ be the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC.$

Let $E, D, F$ be the projection $P$ on sides $AC, BC, AB,$ respectively.

Let $E', D', F'$ be the projection $Q$ on sides $AC, BC, AB,$ respectively.

Prove that points $D, D', E, E', F, F'$ are concyclic.

The midpoint $PQ$ is circumcenter of $DD'EE'FF'.$

Proof

Let $\theta = \angle ACP = \angle BCQ, \Theta = \angle ACQ = \angle BCP.$ $CE \cdot CE' = PC \cos \theta \cdot QC \cos \Theta = PC \cos \Theta \cdot QC \cos \theta = CD \cdot CD'.$

Hence points $D, D', E, E'$ are concyclic.

$PQE'E$ is trapezoid, $E'E \perp PE \implies OE = OE' \implies$

the midpoint $PQ$ is circumcenter of $DD'EE'.$

Similarly points $D, D', F, F'$ are concyclic and points $F, F', E, E'$ are concyclic.

Therefore points $D, D', E, E', F, F'$ are concyclic, so the midpoint $PQ$ is circumcenter of $DD'EE'FF'.$

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Common circumcircle of the pedal triangles as the sign of isogonally conjugate points

Let triangle $\triangle ABC$ and points $P$ and $Q$ inside it be given. Let $D, E, F$ be the projections $P$ on sides $BC, AC, AB,$ respectively. Let $D', E', F'$ be the projections $Q$ on sides $BC, AC, AB,$ respectively.

Let points $D, E, F, D', E', F'$ be concyclic and none of them lies on the sidelines of $\triangle ABC.$

Then point $Q$ is the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC.$

This follows from the uniqueness of the conjugate point and the fact that the line intersects the circle in at most two points.

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Two pares of isogonally conjugate points

Let triangle $\triangle ABC$ and points $X$ and $Y$ be given. Let points $X'$ and $Y'$ be the isogonal conjugate of a points $X$ and $Y$ with respect to a triangle $\triangle ABC,$ respectively.

Let $XY$ cross $X'Y'$ at $Z$ and $XY'$ cross $X'Y$ at $Z'.$

Prove that point $Z'$ is the isogonal conjugate of a point $Z$ with respect to $\triangle ABC.$

Proof

There are two pairs of isogonals $CX - CX'$ and $CY - CY'$ with respect to the angle $\angle ACB \implies$ $CZ - CZ'$ are isogonals with respect to the $\angle ACB$ in accordance with The isogonal theorem.

Similarly $AZ - AZ'$ are the isogonals with respect to the $\angle BAC.$

Therefore the point $Z'$ is the isogonal conjugate of a point $Z$ with respect to $\triangle ABC.$

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Circles

Let $Q$ be the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC.$

Let $D$ be the circumcenter of $\triangle BCP.$

Let $E$ be the circumcenter of $\triangle BCQ.$

Prove that points $D$ and $E$ are inverses with respect to the circumcircle of $\triangle ABC.$

Proof

The circumcenter of $\triangle ABC$ point $O,$ and points $D$ and $E$ lies on the perpendicular bisector of $BC.$ \[\angle BOD = \angle COE = \angle BAC.\] \[2 \angle BDO = \angle BDC = \overset{\Large\frown} {BC} =\] \[= 360^\circ - \overset{\Large\frown} {CB} = 360^\circ - 2 \angle BPC.\] \[\angle BDO = 180^\circ - \angle BPC = \angle PBC + \angle PCB.\] Similarly $\angle CEO = 180^\circ - \angle BQC = \angle QBC + \angle QCB.$ \[\angle PBC + \angle QBC = \angle PBC + \angle PBA = \angle ABC.\] \[\angle QCB + \angle PCB = \angle QCB + \angle QCA = \angle ACB.\]

\[\angle CEO +\angle BDO = \angle ABC + \angle ACB = 180^\circ - \angle BAC \implies\] \[\angle OBD = 180^\circ - \angle BOD - \angle BDO = \angle OEC \implies\] \[\triangle OBD \sim \triangle OEC \implies \frac {OB}{OE} = \frac {OD}{OC} \implies OD \cdot OE = OB^2. \blacksquare\]

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Equidistant isogonal conjugate points

Let triangle $ABC$ with incenter $I$ be given. Denote $\omega = \odot BIC.$

Let point $P'$ be the isogonal conjugate of the point $P$ with respect to $\triangle ABC.$

Prove that $AP = AP'$ iff $P \in \omega.$

Proof

1. Let $P \in \omega.$ WLOG, $P \in \angle BAI.$ Point $P \in \omega \implies \angle PBI = \angle PCI.$

Point $P'$ is the isogonal conjugate of the point $P$ with respect to $\triangle ABC \implies$ \[\angle PBI = \angle P'BI, \angle PCI = \angle P'CI \implies\] \[\angle P'BI = \angle P'CI.\] So points $B,C,I, P,$ and $P'$ are concyclic.

Let $E = AI \cap \odot ABC.$ Then $E$ is the center of $\omega \implies$ \[EP = EP', \angle IEP = \angle IEP' = 2 \angle PBI.\] \[\triangle AEP = \triangle AEP' \implies AP = AP'.\]


2. Let $AP = AP'.$ $\angle PAI = \angle PAE = \angle P'AI = \angle P'AE \implies$

Points $P$ and $P'$ are symmetric with respect $AI \implies PE = P'E.$

Suppose that $P \notin \odot BIC.$

Let $O$ be the center of $\odot BPC, O'$ be the center of $\odot BP'C.$

It is known that points $O$ and $O'$ are inverted with respect to the circumcircle of $\triangle ABC.$

Points $O, O',$ and $E$ belong to bisector $BC, E \in \odot ABC.$

Therefore $\overset{\Large\frown} {BIC}$ divide $\overset{\Large\frown} {BPC}$ and $\overset{\Large\frown} {BP'C}.$

WLOG (see diagram) $PE > IE > P'E,$ contradiction.

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Simplified distance formula for isogonal points

Let triangle $\triangle ABC,$ points $P$ and $P',$ and $\odot ABC = \Omega$ be given. Let point $P'$ be the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC.$ \[D = AP \cap BC, E = AP' \cap BC,\] \[F = AP \cap \Omega, G = AP' \cap \Omega.\] Prove that $PF \cdot P'G= AF \cdot EG.$

Proof

$\angle AFC$ and $\angle AGC$ are both subtended by arc $\overset{\Large\frown} {AC} \implies \angle AFC = \angle AGC.$ \[\angle PCF = \angle PCB + \angle BCF = \angle P'CA + \angle BAF = \angle P'CA + \angle P'AC=\] \[= \angle GP'C \implies \triangle CPF \sim \triangle P'CG \implies PF \cdot P'G = FC \cdot CG.\] Similarly \[\triangle CAF \sim \triangle ECG \implies AF \cdot EG = FC \cdot CG.\] Product of isogonal segments

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Point on circumcircle

Let triangle $\triangle ABC,$ points $D \in BC$ and $E \in BC$ be given.

Denote $\Omega = \odot ABC, \omega = \odot AED,$ \[G = \omega \cap \Omega \ne A, F = AG \cap BC,\] \[K = AE \cap \Omega \ne A, L = GD \cap \Omega \ne G.\] Prove that $KL || BC.$

Proof

WLOG, the order of the points is $B,E,F,C,D,$ as shown on diagram.

The spiral symilarity centered at $A$ maps $\Omega$ to $\omega$ and point $L \in \Omega$ to point $D \in \omega \implies \overset{\Large\frown} {AL} = \overset{\Large\frown} {AD} \implies \angle ABL = \angle AED.$

$\angle AED$ is the external angle of $\triangle AEB \implies \angle AED = \angle ABC + \angle BAE \implies$ \[\angle CBL = \angle BAK \implies \overset{\Large\frown} {BK} = \overset{\Large\frown} {CL} \implies BC ||KL. \blacksquare\]

Corollary

$AL$ is the isogonal conjugate to $AK$ with respect $\angle BAC.$

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Fixed point on circumcircle

Let triangle $\triangle ABC,$ point $G \ne A$ on circumcircle $\Omega = \odot ABC,$ and point $D \in BC$ be given.

Point $P$ lies on $AG,$ point $P'$ be the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC, Q = DP' \cap AP, F = \odot DPQ \cap \Omega.$

Prove that $F$ is fixed point and not depends from position of $P.$

Proof

WLOG, the order of points on sideline is $B, C, D,$ point $B$ is closer to $AP$ than to $AP'.$

Denote $Y = AP \cap BC, Z = AP' \cap BC,\omega = \odot ADY,$ \[F' = \omega \cap \Omega \ne A, H = \Omega \cap F'D \ne F'.\]

Spiral similarity centered at $A$ which maps $\Omega$ into $\odot AYD$ transform point $H$ into point $D \implies$ \[\overset{\Large\frown} {ACH} = \overset{\Large\frown} {AD} \implies\] \[\angle AGH = \angle AYD \implies GH||BC \implies\] \[\overset{\Large\frown} {CH} = \overset{\Large\frown} {BG} \implies\] \[\angle BAG = \angle CAH \implies H \in AP'.\] Points $F', H,$ and $D$ are collinear.

It is known ( Ratio of isogonal segments) that $\frac {AP'}{P'Z} \cdot \frac {AP}{PY} = \frac {AH}{HZ}.$

We use the ratio of the areas and get: \[\frac {[AQD]}{[QYD]} = \frac{AQ}{QY}, \frac {[ZQD]}{[YQD]} = \frac{ZD}{YD},\] \[\frac {[AQD]}{[ZQD]} = \frac{AP'}{P'Z} \implies\] \[\frac{AQ}{QY} = \frac {[AQD]}{[QYD]} = \frac {[AQD]}{[ZQD]} \cdot \frac {[ZQD]}{[YQD]} = \frac{AP'}{P'Z} \cdot \frac{ZD}{YD}.\] Denote $X = AP \cap DH.$ \[\frac {[AXD]}{[YXD]} = \frac{AX}{XY}, \frac {[XZD]}{[XYD]} = \frac{ZD}{YD},\] \[\frac {[AXD]}{[XZD]} = \frac{AH}{HZ} \implies\] \[\frac{AX}{XY} = \frac {[AXD]}{[YXD]} = \frac {[AXD]}{[XZD]} \cdot \frac {[XZD]}{[YXD]} = \frac{AH}{HZ} \cdot \frac{ZD}{YD}.\] Therefore $\frac{AX}{XY} = \frac{AQ}{QY} \cdot \frac{AP}{PY}$ which means ( Problems | Simple) that $DX$ is the radical axes of $\omega$ and $\odot DPQ \implies$

$F = F'$ and not depends from position of $P.$

Fixed point on circumcircle

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Distance formula for isogonal points

Let triangle $\triangle ABC$ and point $P$ be given.

Let point $P'$ be the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC.$

Let lines $AP$ and $AP'$ cross sideline $BC$ at $D$ and $E$ and circumcircle of $\triangle ABC$ at $F$ and $G,$ respectively.

We apply the Isogonal’s property and get $\frac {BD}{DC} \cdot \frac{BE}{EC} = \frac {AB^2}{AC^2}.$

$EF || BC.$ We apply the Ptolemy's theorem to $BCGF$ and get \[BC \cdot FG = BG^2 – BF^2.\]

We apply the barycentric coordinates and get \[\left| \frac{BE}{EC} - \frac {BD}{DC} \right| = \frac {AB \cdot BC \cdot FG}{AC \cdot PF \cdot P'G}.\]

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Miquel point for isogonal conjugate points

Let triangle $\triangle ABC,$ points $Q \in BC$ and $P$ be given. Let point $P'$ be the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC.$ \[D = QP \cap AP', E = AP \cap QP'.\] Let $M$ be the Miquel point of a complete quadrilateral $PDP'E.$

Prove that $M$ lies on the circumcircle of $\triangle ABC.$

Proof

Point $A$ is the isogonal conjugate of a point $Q$ with respect to a triangle $\triangle ABC,$ so point $D$ is the isogonal conjugate of a point $E$ with respect to a triangle $\triangle ABC.$

Points $P$ and $E$ lies on the same line, therefore \[PF \cdot P'G = EF \cdot DG \implies\]

\[\frac {PF-EP}{PF} = \frac{DG-DP'}{DG} \implies\] \[\frac {EP}{PF} = \frac{DP'}{DG}.\] Point $M$ lies on circles $APD$ and $AEP' \implies$ spiral similarity centered at $M$ transform triangle $\triangle MPE$ to $\triangle MDP' \implies$ \[\angle MPE = \angle MDP', \frac {MP}{MD} = \frac {PE}{DP'} = \frac{PF}{DG} \implies\] \[\triangle MPF \sim \triangle MDG \implies\] \[\angle AFM = \angle AGM \implies M \in \odot ABC.\]

vladimir.shelomovskii@gmail.com, vvsss

Point on circumcircle

Let triangle $\triangle ABC,$ and points $D \in BC$ and $E \in \odot ABC = \Omega$ be given.

Let $\odot ADE = \omega, F = BC \cap \omega \ne D.$

Let lines $AF$ and $AG (G \in BC)$ be the isogonals with respect to the angle $\angle BAC, \odot AGD = \theta.$

Let $P$ be an arbitrary point on $AF, Q = DP \cap AG, H = \theta \cap \Omega.$

Prove that $X = EP \cap HQ$ lies on $\Omega.$

Simplified problem

Let $\triangle ABC,$ and points $D \in BC$ and $E \in \odot ABC = \Omega$ be given, $\omega = \odot ADE, F = \omega \cap BC \ne D.$

Let lines $AF$ and $AG (G \in BC)$ be the isogonals with respect to $\angle BAC, \theta = \odot AGD, H = \theta \cap \Omega.$

Prove that $X = EF \cap HG \in \Omega.$

Proof, Simplified problem \[\angle CGH = \angle DAH = \frac{1}{2} \overset{\Large\frown} {HD} (\theta),\] \[\angle DFE = \angle DAE = \frac{1}{2} \overset{\Large\frown} {ED} (\omega),\] \[\angle EAH = \angle DAE - \angle DAH = \frac{1}{2} \overset{\Large\frown} {EH}(\Omega),\] \[\angle EXH = \angle DFE - \angle DGH = \angle EAH \implies\]

points $A, H, E, X$ are concyclic on $\Omega.$

Proof

Let points $P'$ and $Q'$ be the isogonal conjugate of a points $P$ and $Q$ with respect to a triangle $\triangle ABC, \omega' = \odot Q'PD, \theta' = \odot P'QD.$

It is known that $E \in \omega', H \in \theta', \omega' \cap \theta' \cap \Omega = M.$

\[\angle DQH = \angle DMH = \frac{1}{2} \overset{\Large\frown} {HD} (\theta'),\] \[\angle DPE = \angle DME = \frac{1}{2} \overset{\Large\frown} {ED} (\omega'),\] \[\angle EMH = \angle DME - \angle DMH = \frac{1}{2} \overset{\Large\frown} {EH}(\Omega),\] \[\angle EXH = \angle DPE - \angle DQH = \angle EMH \implies\] points $M, H, E, X$ are concyclic on $\Omega.$

vladimir.shelomovskii@gmail.com, vvsss

Isogonal of line BC with respect to angle BAC

Let triangle $\triangle ABC$ be given, $\Omega = \odot ABC, AD || BC.$

Let lines $AE$ and $AD$ be the isogonals with respect to $\angle BAC.$

Prove that $AE$ is tangent to $\Omega.$

Proof

Let $O$ and $H$ be the circumcenter and the orthocenter of $\triangle ABC,$ respectively. \[AH \perp BC, AD || BC \implies AH \perp AD.\] $AH$ is isogonal to $AO$ with respect to $\angle BAC \implies AE \perp AO \implies AE$ is tangent to $\Omega.$

vladimir.shelomovskii@gmail.com, vvsss

Isogonal bijection lines and points

Let triangle $\triangle ABC$ and line $\ell, P \in \ell$ be given, $\Omega = \odot ABC.$

Define $G \in \Omega$ the point with property $G' \in \ell.$

Prove that $\angle ABG$ is equal the angle $\theta$ between $\ell$ and $BC.$

Proof

WLOG, the configuration is the same as shown on diagram, $F = \ell \cap BC, AD' || \ell, \theta = \angle PFB, AD || BC,$ $AE$ is the tangent to $\Omega.$

$AD$ is isogonal to $AE, AD'$ is isogonal to $AG$ with respect to $\angle BAC \implies$ \[\theta = \angle PFB = \angle D'AD = \angle GAE = \angle GBA.\] A bijection has been established between the set of lines parallel to a given one and the set of points of the circumcircle.

vladimir.shelomovskii@gmail.com, vvsss

Miquel point for two pare isogonal points

Let triangle $\triangle ABC$ and points $P$ and $Q$ be given.

Let points $P'$ and $Q'$ be the isogonal conjugate of the points $P$ and $Q$ with respect to $\triangle ABC, \Omega = \odot ABC, M$ is the Miquel point of quadrilateral $PQP'Q'.$

Prove that $M \in \Omega.$

Proof

Denote $R = PQ \cap P'Q', \Theta = \odot P'QR, \theta = \odot PQ'R.$

Then $M = \theta \cap \Theta$ is the Miquel point of quadrilateral $PQP'Q'.$

Denote $E = \theta \cap \Omega \notin \Theta, F = \Theta \cap \Omega \notin \theta.$

Let $D \in \Omega$ be the point with property $D' \in PQ.$

WLOG, configuration is similar as shown in diagram.

$P' \in DF, Q' \in DE$ ( Isogonal_bijection_lines_and_points). \[\angle EMF = \angle RME - \angle RMF = \angle RQ'E - \angle RP'F = \angle P'Q'E - \angle DP'Q' = \angle P'DQ' = \angle EDF \implies M \in \odot DEF \blacksquare\]

vladimir.shelomovskii@gmail.com, vvsss

Isogonic center’s conjugate point

Let triangle $ABC$ with isogonic center $F (X(13)$ or $X(14))$ be given. Denote $\omega = \odot BIC.$

Let line $\ell_A$ be the axial symmetry of line $AF$ according to the sideline $BC.$

Define lines $\ell_B$ and $\ell_C$ similarly.

Prove that the lines $\ell_A, \ell_B,$ and $\ell_C$ are concurrent.

Proof

Let $I$ be the incenter of $\triangle ABC, F =X(13).$ \[A_1 = AF \cup BC, E = AI \cup BC, D \in BC,\] \[AD \perp AE, \omega = \odot AED.\] Let $F_1 = AF \cup \omega, F'$ is simmetric to $F_1$ with respect $BC \implies A_1F' = \ell_A.$

The diameter $DE$ of $\omega$ lies on $BC \implies F_1 \in \omega, \overset{\Large\frown} {EF_1} = \overset{\Large\frown} {EF'} \implies \angle EAF_1 = \angle EAF'.$

Therefore $AF'$ is the isogonal conjugate of $AF$ with respect to $\angle BAC.$

Similarly $\ell_B$ and $\ell_C$ are the isogonal conjugate of $BF$ and $CF,$ so point $F'$ is the isogonal conjugate of point $F$ with respect to $\triangle ABC.$

The second diagram show construction in the case $F =X(14).$ The proof is similar.

vladimir.shelomovskii@gmail.com, vvsss

Three pairs isogonal points

Let a triangle $ABC,$ points $D$ and $E \in AD$ be given, $F = CD \cap BE.$ Points $D', E'$ and $F'$ are the isogonal conjugate of the points $D, E,$ and $F,$ respectively, with respect to $\triangle ABC.$

Prove that $\frac {AD}{AE} \cdot \frac {BE}{BF} \cdot \frac {CF}{CD} = \frac {AE'}{AD'} \cdot \frac {BF'}{BE'} \cdot \frac {CD'}{CF'}.$

Proof

Denote $\angle BAC = \alpha, \angle ABC = \beta, \angle ACB = \gamma,$ \[\angle BAD = \varphi_A, \angle CBE = \varphi_B, \angle ACD = \varphi_C.\] We use isogonal properties and get \[\angle CAD' = \varphi_A, \angle ABE' = \varphi_B, \angle BCD' = \varphi_C.\] By applying the Law of Sines, we get \[\frac {BE}{AE} = \frac {\sin \varphi_A}{\sin (\beta - \varphi_B)}, \frac {CF}{BF} = \frac {\sin \varphi_B}{\sin (\gamma - \varphi_C)}, \frac {AD}{CD} = \frac {\sin \varphi_C}{\sin (\alpha - \varphi_A)}.\] Symilarly, \[\frac {AE'}{BE'} = \frac {\sin \varphi_B}{\sin (\alpha - \varphi_A)}, \frac {BF'}{CF'} = \frac {\sin \varphi_C}{\sin (\beta - \varphi_B)}, \frac {CD'}{AD'} = \frac {\sin \varphi_A}{\sin (\gamma - \varphi_C)}.\] We multiply these equations and get \[\frac {AE \cdot BF \cdot CD}{AD \cdot BE \cdot CF} = \frac{AD' \cdot BE' \cdot CF'}{AE' \cdot BF' \cdot CD'} = \frac {\sin \varphi_A \cdot \sin \varphi_B \cdot \sin \varphi_C}{\sin (\alpha - \varphi_A) \cdot \sin (\beta - \varphi_B) \cdot \sin (\gamma - \varphi_C)}.\] vladimir.shelomovskii@gmail.com, vvsss

Ratio for three pairs of isogonal points

Let a triangle $ABC,$ points $D$ and $E \in AD$ be given, $F = CD \cap BE.$

Points $D', E'$ and $F'$ are the isogonal conjugate of the points $D, E,$ and $F,$ respectively, with respect to $\triangle ABC.$

Denote $R$ and $R'$ the circumradii of triangles $\triangle DEF$ and $\triangle D'E'F',$ respectively.

Prove that $\frac {AD \cdot BE \cdot CF}{R} = \frac {AE' \cdot BF' \cdot CD'}{R'}.$

Proof

Denote $\frac {AD}{DE} = u, \frac {BE}{EF} = w, \frac {CF}{FD} = v.$

$[DEF] = 1,$ where $[X]$ is the area of the figure $X.$ \[\frac {[ADF]}{[DEF]} = \frac {AD}{DE} = u,\] \[\frac {[ACF]}{[DEF]} = \frac {[ACF]}{[ADF]} \cdot \frac {[ADF]}{[DEF]} = uv.\] Similarly, $\frac {[CFE]}{[DEF]} = v, \frac {[BCE]}{[DEF]} = vw,$ \[\frac {[BDE]}{[DEF]} = w, \frac {[ABD]}{[DEF]} = uw.\] \[\frac {[ABC]}{[DEF]} = 1 + u + v + w + uv + uw + vw = (1 + u)(1 + v)(1 + w) - uvw,\] \[\frac {[ABC]}{[DEF]} = \frac {AE \cdot BF \cdot CD}{DE \cdot DF \cdot EF} - \frac {AD \cdot BE \cdot CF}{DE \cdot DF \cdot EF}.\] \[\frac {DE \cdot DF \cdot EF}{4R} = [DEF] \implies\] \[[ABC] = \frac {AD \cdot BE \cdot CF}{4R} \cdot \left ( \frac {AE \cdot BF \cdot CD}{BE \cdot CF \cdot AD}-1 \right).\] Similarly, \[[ABC] = \frac {AE' \cdot BF' \cdot CD'}{4R'} \cdot \left ( \frac {AD' \cdot BE' \cdot CD}{AE' \cdot BF' \cdot CD'}-1 \right).\] It is known that $\frac {AD}{AE} \cdot \frac {BE}{BF} \cdot \frac {CF}{CD} = \frac {AE'}{AD'} \cdot \frac {BF'}{BE'} \cdot \frac {CD'}{CF'}$ ( Three pairs isogonal points), therefore \[\frac {AD \cdot BE \cdot CF}{4R} = \frac {AE' \cdot BF' \cdot CD'}{4R'}.\] Comment: The main idea of the proof was found by Leonid Shatunov.

vladimir.shelomovskii@gmail.com, vvsss

Problems

  • Given a nonisosceles, nonright triangle $ABC,$ let $O$ denote the center of its circumscribed circle, and let $A_1, \, B_1,$ and $C_1$ be the midpoints of sides $BC, \, CA,$ and $AB,$ respectively. Point $A_2$ is located on the ray $OA_1$ so that $\triangle OAA_1$ is similar to $\triangle OA_2A$. Points $B_2$ and $C_2$ on rays $OB_1$ and $OC_1,$ respectively, are defined similarly. Prove that lines $AA_2, \, BB_2,$ and $CC_2$ are concurrent. (Source)
  • Let $P$ be a given point inside quadrilateral $ABCD$. Points $Q_1$ and $Q_2$ are located within $ABCD$ such that $\angle Q_1 BC = \angle ABP$, $\angle Q_1 CB = \angle DCP$, $\angle Q_2 AD = \angle BAP$, $\angle Q_2 DA = \angle CDP$. Prove that $\overline{Q_1 Q_2} \parallel \overline{AB}$ if and only if $\overline{Q_1 Q_2} \parallel \overline{CD}$. (Source)
  • Let $AA', BB',$ and $CC'$ be the bisectors of a triangle $\triangle ABC.$
The segments $BB'$ and $A'C'$ meet at point $D.$ Let $E$ be the projection of $D$ to $AC.$
Points $P$ and $Q$ on the sides $AB$ and $BC,$ respectively, are such that $EP = PD, EQ = QD.$
Prove that $\angle PDB' = \angle EDQ.$ (Source)
  • IMO 2007 Short list/G3

The diagonals of a trapezoid $ABCD$ intersect at point $P.$

Point $Q$ lies between the parallel lines $BC$ and $AD$ such that $\angle AQD = \angle CQB,$ and line $CD$ separates points $P$ and $Q.$

Prove that $\angle BQP = \angle DAQ.$

Proof

$\angle AQD = \angle CQB \implies$

$BQ$ and $AQ$ are isogonals with respect $\angle CQD.$

$P =AC \cap BD, BC || AD \implies$

$QS || AD$ is isogonal to $QP$ with respect $\angle CQD.$

From the converse of The isogonal theorem we get

$\angle BQP = \angle SQA = \angle DAQ \blacksquare$

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