# Isogonal conjugate

Isogonal conjugates are pairs of points in the plane with respect to a certain triangle.

## 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. $\blacksquare$

## 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 $AH \perp BC.$ $\blacksquare$

## 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 $\angle BDE = \angle FDC.$

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

Proof

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

$\angle BDE = \angle FDC \implies DE$ 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 $DG.$ $\blacksquare$

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

## 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 $\angle EPB = \angle FPD.$

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 $PF. \blacksquare$

## 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 $MD. \blacksquare$

## 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 $(\ell,A). \blacksquare$

## 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 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. $\blacksquare$

## 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

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

## Definition of isogonal conjugate of a point

Let $P$ be a point in the plane, and let $ABC$ be a triangle. We will denote by $a,b,c$ the lines $BC, CA, AB$. Let $p_a, p_b, p_c$ denote the lines $PA$, $PB$, $PC$, respectively. Let $q_a$, $q_b$, $q_c$ be the reflections of $p_a$, $p_b$, $p_c$ over the angle bisectors of angles $A$, $B$, $C$, respectively. Then lines $q_a$, $q_b$, $q_c$ concur at a point $Q$, 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 $\angle A.$

## 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 $\angle BAC.$ $\blacksquare$

## 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 $$PC = P_1C, PC = P_2C \implies P_1C = P_2C.$$ $$\angle ACQ = \angle BCP_1 \implies \angle QCP_1 = \angle ACB.$$ $$\angle BCQ = \angle ACP_2 \implies \angle QCP_2 = \angle ACB.$$ $\angle QCP_1 = \angle QCP_2, CP_1 = CP_2, QC$ common $\implies$ $$\triangle QCP_1 = \triangle QCP_2 \implies QP_1 = QP_2.$$ 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 $$(p' : q' : r'), p' = \frac {|B - C|^2}{p}, q' = \frac {|A-C|^2}{q}, r' = \frac {|A - B|^2}{r}.$$

## 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 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 $\angle ABP = \angle CBQ \implies$ point $Q$ is the isogonal conjugate of a point $P$ with respect to a triangle $\triangle ABC. \blacksquare$

## 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.

Then 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 $DD'EE'FF'.$

## 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.

## 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 $\angle CEO = 180^\circ - \angle BQC = \angle QBC + \angle QCB.$ $$\angle PBC + \angle QBC = \angle PBC + \angle PBA = \angle ABC.$$ $$\angle QCB + \angle PCB = \angle QCB + \angle QCA = \angle ACB.$$

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

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, i.e. these three lines intersect at a point. (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)