On Some Properties of a Complete Quadrangle

Abstract

In this paper, we study the properties of a complete quadrangle in the Euclidean plane. The proofs are based on using rectangular coordinates symmetrically on four vertices and four parameters $a,b,c,d$. Here, many properties of the complete quadrangle known from earlier research are proved using the same method, and some new results are given.

1. Introduction

If four points are joined in pairs by six distinct lines, they are called the vertices of a complete quadrangle, and the lines are its six sides. Two sides are said to be opposite if they have no common vertex. The study of the geometry of the complete quadrangle has a long history and there are numerous articles in which the properties of quadrangles have been studied. In this paper, we deal with the properties of quadrangles related to the center and anticenter of the quadrangle, the diagonal triangle of the quadrangle, and isogonality with respect to the four triangles formed by the vertices of the quadrangle. These properties were studied in the literature using a number of various methods [1,2,3,4,5,6,7,8,9,10,11,12].

Our approach in this paper uses a novel method which is applicable to studying and extending the known properties of a quadrangle. We put the complete quadrangle into such a coordinate system that its circumscribed hyperbola is rectangular. We use this method to prove the 12 theorems already published in aforementioned papers and to derive two new original theorems, Theorems 8 and 14, which to our knowledge were not yet published in the literature. Thus, our method allows one to study the properties of quadrangles in a more unified way.

In our former work in [13], we analyzed a complete quadrilateral in a similar way. A complete quadrilateral is a set of four lines (sides of the quadrilateral), where none of two lines are parallel and none of the three are concurrent. Using the fact that a unique parabola can be inscribed on each quadrilateral, the coordinate system was chosen so that the parabola has the equation $y^2=4x$. The sides of the quadrilateral are given by $ay=x+a^2$, $by=x+b^2$, $cy=x+c^2$, $dy=x+d^2$ where $a,b,c,d$ are real numbers. The coordinate system chosen in this way is suitable for studying quadrilaterals, but not for studying quadrangles.

As in [14], we proved:

Lemma 1. 

For each quadrangle for which the opposite sides are not perpendicular, the rectangular hyperbola can be circumscribed.

Therefore, we choose the coordinate system for studying complete quadrangles such that circumscribed hyperbola of the complete quadrangle is given by $xy=1$. In the same paper, we studied the quadruples of orthopoles.

2. Methods

Let $ABCD$ be a complete quadrangle and $\mathcal{H}$ be a rectangular hyperbola circumscribed to it. With the suitable choice of the coordinate system, it can be achieved that $\mathcal{H}$ has the equation $xy=1$ and the vertices of the quadrangle are of the form

$$A=\left(a,\frac{1}{a}\right),B=\left(b,\frac{1}{b}\right),C=\left(c,\frac{1}{c}\right),D=\left(d,\frac{1}{d}\right),$$

(1)

where $a,b,c,d\ne 0$.

Let $s,q,r,p$ be elementary symmetric functions in four variables $a,b,c,d$:

$$\begin{array}{c}\hfill s=a+b+c+d,q=ab+ac+ad+bc+bd+cd,\\ \hfill r=abc+abd+acd+bcd,p=abcd.\end{array}$$

The centroid of the quadrangle $ABCD$ is of the form

$$G=\left(\frac{s}{4},\frac{r}{4p}\right).$$

(2)

The sides of $ABCD$ have the equations:

$$\begin{array}{c}\hfill AB\cdots x+aby=a+b,AC\cdots x+acy=a+c,AD\cdots x+ady=a+d\\ \hfill BC\cdots x+bcy=b+c,BD\cdots x+bdy=b+d,CD\cdots x+cdy=c+d.\end{array}$$

(3)

The choice of the equation of the hyperbola $\mathcal{H}$, i.e., the coordinates of the vertices, enables us to prove the claims in a simple way using an analytical method. The calculations are elementary and mostly very short.

The paper is organized in such a way that we first prove a property, and then state it in a theorem. After the theorem, we point out whether the claim is previously known from the literature or is our original contribution.

3. Results

3.1. The Center and Anticenter of the Quadrangle $ABCD$

In this section, we study the Euler circles of four triangles of the quadrangle $ABCD$, and define its center and anticenter. The circle with the equation

$$2abc\left(x^2+y^2\right)+\left[1-abc\left(a+b+c\right)\right]x-\left(a^2{b}^2{c}^2-ab-ac-bc\right)y=0$$

passes through the midpoint $(\frac{1}{2}(a+b),\frac{1}{2ab}(a+b))$ of points A and B. Similarly, it passes through the midpoints of $A,C$, i.e., $B,C$, so it is Euler’s circle ${\mathcal{N}}_d$ of the triangle $ABC$. It obviously passes through the origin O. Analogously, the same is valid for Euler’s circles ${\mathcal{N}}_c$, ${\mathcal{N}}_b$, and ${\mathcal{N}}_a$ of the triangles $ABD$, $ACD$, and $BCD$. Hence, we have just proved the following statement:

Theorem 1. 

Euler’s circles of the triangles $BCD$, $ACD$, $ABD$, and $ABC$ of the complete quadrangle with the circumscribed rectangular hyperbola passes through the center of the hyperbola.

The theorem is coming from [3].

There are several names for the point O in the literature. In this paper, we call it the center of the quadrangle $ABCD$. The point $O^{\prime }=(\frac{s}{2},\frac{r}{2p})$, symmetric to the point O with respect to the centroid G, we call the anticenter of the quadrangle $ABCD$. The asymptotes $\mathcal{X}$ and $\mathcal{Y}$ of the hyperbola $\mathcal{H}$ are the axes of the quadrangle $ABCD$.

The center $N_d$ of the circle ${\mathcal{N}}_d$, i.e., Euler’s center of the triangle $ABC$, is the point

$$N_d=\left(\frac{1}{4}\left(a+b+c-\frac{1}{abc}\right),\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-abc\right)\right).$$

(4)

The Euler’s centers $N_a,N_b,N_c$ of the triangles $BCD$, $ACD$, $ABD$ are of similar forms. The distance from $N_d$ to the origin O fulfills

$$\begin{array}{ccc}\hfill O{N_d}^2& =& {\left(\frac{1}{4abc}\right)}^2{\left[abc(a+b+c)-1\right]}^2+{(ab+ac+bc-a^2{b}^2{c}^2)}^2\hfill \\ & =& {\left(\frac{1}{4abc}\right)}^2(a^2{b}^2+1)(a^2{c}^2+1)(b^2{c}^2+1).\hfill \end{array}$$

Hence, Euler’s circle of the triangle $ABC$ has the radius $\frac{1}{4}\left|\frac{d}{p}\right|\sqrt{(a^2{b}^2+1)(a^2{c}^2+1)(b^2{c}^2+1)}$. The other three radii of the Euler’s circles of the other three triangles look quite similar, as can be seen in [11]. Because of that, the radius ${\rho }_d$ of the circumscribed circle of the triangle $ABC$ is given within following analogous formulae

$${\rho }_a=\frac{1}{2}\left|\frac{a}{p}\right|\sqrt{{\lambda }^{\prime }{\mu }^{\prime }{\nu }^{\prime }},{\rho }_b=\frac{1}{2}\left|\frac{b}{p}\right|\sqrt{{\lambda }^{\prime }\mu \nu },{\rho }_c=\frac{1}{2}\left|\frac{c}{p}\right|\sqrt{\lambda {\mu }^{\prime }\nu },{\rho }_d=\frac{1}{2}\left|\frac{d}{p}\right|\sqrt{\lambda \mu {\nu }^{\prime }},$$

where ${\rho }_a,{\rho }_b,{\rho }_c$ are the radii of the circumscribed circles of the triangles $BCD$, $ACD$, $ABD$ using the following notations

$$\begin{array}{ccc}\hfill \lambda =a^2{b}^2+1,& \mu =a^2{c}^2+1,& \nu =a^2{d}^2+1,\hfill \\ \hfill {\lambda }^{\prime }=c^2{d}^2+1,& {\mu }^{\prime }=b^2{d}^2+1,& {\nu }^{\prime }=b^2{c}^2+1,\hfill \end{array}$$

(5)

where $\lambda ,\mu ,\nu ,{\lambda }^{\prime },{\mu }^{\prime },{\nu }^{\prime }>0$. The parameters (5) appear in formulae for the lengths of the sides of the quadrangle $ABCD$. Indeed, for the points A and B, we obtain

$${AB}^2={(a-b)}^2+{\left(\frac{1}{a}-\frac{1}{b}\right)}^2={\left(\frac{a-b}{ab}\right)}^2(a^2{b}^2+1)={\left(\frac{a-b}{ab}\right)}^2\lambda ,$$

i.e., $AB=\left|\frac{a-b}{ab}\right|\sqrt{\lambda }$. The other five analogous statements are also valid

$$AC=\left|\frac{a-c}{ac}\right|\sqrt{\mu },AD=\left|\frac{a-d}{ad}\right|\sqrt{\nu },BC=\left|\frac{b-c}{bc}\right|\sqrt{{\nu }^{\prime }},BD=\left|\frac{b-d}{bd}\right|\sqrt{{\mu }^{\prime }},CD=\left|\frac{c-d}{cd}\right|\sqrt{{\lambda }^{\prime }}.$$

From these equalities, the next equalities follow

$$\begin{array}{c}\hfill AB·CD=\left|\frac{(a-b)(c-d)}{p}\right|\sqrt{\lambda {\lambda }^{\prime }},AC·BD=\left|\frac{(a-c)(b-d)}{p}\right|\sqrt{\mu {\mu }^{\prime }},\\ \hfill AD·BC=\left|\frac{(a-d)(b-c)}{p}\right|\sqrt{\nu {\nu }^{\prime }}.\end{array}$$

For the coordinates of the point $N_d$ from (4), it proves that

$$\left(x-\frac{s}{4}\right)\left(y-\frac{r}{4p}\right)=\frac{1}{16p}{(p+1)}^2.$$

The same is also valid for $N_a,N_b,N_c$. Therefore, we have proved the result:

Theorem 2. 

The centroid G of the quadrangle $ABCD$ is the center of the quadrangle $N_a{N}_b{N}_c{N}_d$, where $N_a,N_b,N_c,N_d$ are the centers of Euler circles $BCD$, $ACD$, $ABD$, $ABC$, respectively, and the quadrangles $ABCD$ and $N_a{N}_b{N}_c{N}_d$ have the parallel axes.

This result is coming from [7,11]. Because the midpoints $AD,BD,CD$ are symmetric to the midpoints $BC,AC,AB$ with respect to the centroid G, the circle incident to the midpoints of $AD,BD,CD$ is symmetric to the Euler circle ${\mathcal{N}}_d$ of the triangle $ABC$ with respect to the centroid G. Hence, that circle is incident to anticenter $O^{\prime }$ because the circle ${\mathcal{N}}_d$ is incident to O. We have proved the following:

Theorem 3. 

Circles incident to the midpoints of three sides $AD,BD,CD$; $AC,BC,CD$; $AB,BC,BD$; $AB,AC,AD$ are passing through the anticenter $O^{\prime }$.

The result is also given in [1,10].

The line $AB$ has the slope $-\frac{1}{ab}$, and the connecting line of the origin and the midpoint of $AB$ has the slope $\frac{1}{ab}$, so these lines are antiparallel with respect to the coordinate axes. The same is valid for any side of the quadrangle $ABCD$. We showed the result:

Theorem 4. 

The angle of any two sides of the quadrangle is opposite to the angle of the connecting lines of the midpoints of these sides and the center of $ABCD$.

This result was also given in [3,12]. Let us study the points

$$H_a=\left(-\frac{1}{bcd},-bcd\right),H_b=\left(-\frac{1}{acd},-acd\right),H_c=\left(-\frac{1}{abd},-abd\right),H_d=\left(-\frac{1}{abc},-abc\right).$$

(6)

The line with the equation $abx-y=abc-\frac{1}{c}$ is perpendicular to the line $AB$ from (3) and it is incident to C and $H_d$, so the line $C{H}_d$ is height from C of the triangle $ABC$. Similarly, the lines $A{H}_d$ and $B{H}_d$ are the heights from the vertices A and B of the triangle $ABC$. Therefore, $H_d$ is the orthocenter of that triangle. Hence, we showed that the following is valid:

Theorem 5. 

The orthocenters $H_a,H_b,H_c,H_d$ of the triangles $BCD,ACD,ABD,ABC$, respectively, are incident to the rectangular hyperbola $\mathcal{H}$.

This statement has been proven in [3], and it also proves the converse of Lemma 2 from [14].

As the orthocenters $H_a,H_b,H_c,H_d$ are incident to hyperbola $\mathcal{H}$, its center O is the center of the quadrangle $H_a{H}_b{H}_c{H}_d$. Thus, we have proved:

Theorem 6. 

Quadrangles $ABCD$ and $H_a{H}_b{H}_c{H}_d$ have the same center.

This result also appears in [11].

If the point D coincides with $H_d$, then $d=\frac{1}{abc}$, $p=-1$, and the quadrangle $ABCD$ is the orthocentric quadrangle (see [14]).

3.2. A Diagonal Triangle of the Quadrangle $ABCD$

Diagonal points $U=AB\cap CD$, $V=AC\cap BD$, $W=AD\cap BC$ of the quadrangle $ABCD$ are given by

$$U=\left(\frac{ab(c+d)-cd(a+b)}{ab-cd},\frac{a+b-c-d}{ab-cd}\right),V=\left(\frac{ac(b+d)-bd(a+c)}{ac-bd},\frac{a+c-b-d}{ac-bd}\right),$$

$$W=\left(\frac{ad(b+c)-bc(a+d)}{ad-bc},\frac{a+d-b-c}{ad-bc}\right).$$

These points can be written in the shorter form

$$U=\left(\frac{u^{\prime }}{u},\frac{u^{″}}{u}\right),V=\left(\frac{v^{\prime }}{v},\frac{v^{″}}{v}\right),W=\left(\frac{w^{\prime }}{w},\frac{w^{″}}{w}\right),$$

where

$$\begin{array}{ccc}\hfill u=ab-cd,& u^{\prime }=ab(c+d)-cd(a+b),& u^{″}=a+b-c-d,\hfill \\ \hfill v=ac-bd,& v^{\prime }=ac(b+d)-bd(a+c),& v^{″}=a+c-b-d,\hfill \\ \hfill w=ad-bc,& w^{\prime }=ad(b+c)-bc(a+d),& w^{″}=a+d-b-c.\hfill \end{array}$$

The following equalities are valid

$$u^{\prime }{v}^{″}+u^{″}{v}^{\prime }=2uv,u^{\prime }{w}^{″}+u^{″}{w}^{\prime }=2uw,v^{\prime }{w}^{″}+v^{″}{w}^{\prime }=2vw.$$

Therefore, the lines $\mathcal{U}$, $\mathcal{V}$, $\mathcal{W}$ with equations

$$u^{″}x+u^{\prime }y=2u,v^{″}x+v^{\prime }y=2v,w^{″}x+w^{\prime }y=2w$$

are incident to the pairs of points $V,W$; $U,W$; $U,V$, respectively. So, they are the diagonals of the quadrangle $ABCD$. Hence, their equations are

$$\begin{array}{ccc}\hfill \mathcal{U}& \dots & (a+b-c-d)x+\left[ab\right(c+d)-cd(a+b\left)\right]y=2(ab-cd),\hfill \\ \hfill \mathcal{V}& \dots & (a+c-b-d)x+\left[ac\right(b+d)-bd(a+c\left)\right]y=2(ac-bd),\hfill \\ \hfill \mathcal{W}& \dots & (a+d-b-c)x+\left[ad\right(b+c)-bc(a+d\left)\right]y=2(ad-bc).\hfill \end{array}$$

The centroid $G_{UVW}$ of the triangle $UVW$ is the point

$$G_{UVW}=\left(\frac{u^{\prime }vw+u{v}^{\prime }w+uv{w}^{\prime }}{3uvw},\frac{u^{″}vw+u{v}^{″}w+uv{w}^{″}}{3uvw}\right).$$

The heights from vertices U and V of the diagonal triangle $UVW$ have the equations

$$u{u}^{\prime }x-u{u}^{″}y={u^{\prime }}^2-{u^{″}}^2,v{v}^{\prime }x-v{v}^{″}y={v^{\prime }}^2-{v^{″}}^2.$$

For their intersection point $(x,y)$, the equalities

$$\begin{array}{c}\hfill uv(u^{\prime }{v}^{″}-u^{″}{v}^{\prime })x={u^{\prime }}^{2}v{v}^{″}-u{u}^{″}{v^{\prime }}^2+u^{″}{v}^{″}(u{v}^{″}-u^{″}v),\\ \hfill uv(u^{\prime }{v}^{″}-u^{″}{v}^{\prime })y=u^{\prime }{v}^{\prime }(u^{\prime }v-u{v}^{\prime })+u{u}^{\prime }{v^{\prime \prime }}^2-{u^{\prime \prime }}^{2}v{v}^{\prime }\end{array}$$

are valid. However, it can be checked that

$$\begin{array}{ccc}\hfill u^{\prime }{v}^{″}-u^{″}{v}^{\prime }& =& 2(a-d)(b-c)w,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill {u^{\prime }}^{2}v{v}^{″}-u{u}^{″}{v^{\prime }}^2& =& (a-d)(b-c)(u^{\prime }vw+u{v}^{\prime }w+uv{w}^{\prime }),\hfill \end{array}$$

(7)

$$\begin{array}{ccc}\hfill u{v}^{″}-u^{″}v& =& (a-d)(b-c)w^{″},\hfill \end{array}$$

(8)

$$\begin{array}{ccc}\hfill u^{\prime }v-u{v}^{\prime }& =& (a-d)(b-c)w^{\prime },\hfill \\ \hfill u{u}^{\prime }{v^{\prime \prime }}^2-{u^{\prime \prime }}^{2}v{v}^{\prime }& =& (a-d)(b-c)(u^{″}vw+u{v}^{″}w+uv{w}^{″})\hfill \end{array}$$

(9)

are valid. Hence, the orthocenter of the triangle $UVW$ is the point

$$H_{UVW}=\left(\frac{u^{\prime }vw+u{v}^{\prime }w+uv{w}^{\prime }+u^{″}{v}^{″}{w}^{″}}{2uvw},\frac{u^{″}vw+u{v}^{″}w+uv{w}^{″}+u^{\prime }{v}^{\prime }{w}^{\prime }}{2uw}\right).$$

The centroid, orthocenter, and circumcenter $O_{UVW}$ of the triangle $UVW$ fulfill the equality $2O_{UVW}+H_{UVW}=3G_{UVW}$, out of which we obtain

$$O_{UVW}=\left(\frac{u^{\prime }vw+u{v}^{\prime }w+uv{w}^{\prime }-u^{″}{v}^{″}{w}^{″}}{4uvw},\frac{u^{″}vw+u{v}^{″}w+uv{w}^{″}-u^{\prime }{v}^{\prime }{w}^{\prime }}{4uvw}\right).$$

Now let us study the circle ${\mathcal{K}}_{UVW}$ with the center $O_{UVW}$ and the equation

$$2uvw(x^2+y^2)-(u^{\prime }vw+u{v}^{\prime }w+uv{w}^{\prime }-u^{″}{v}^{″}{w}^{″})x-(u^{″}vw+u{v}^{″}w+uv{w}^{″}-u^{\prime }{v}^{\prime }{w}^{\prime })y=0.$$

(10)

We will show the circumscribed circle of the triangle $UVW$. We will also show that U is incident to this circle, and it is proved by the equality

$$2vw({u^{\prime }}^2+{u^{″}}^2)-(u^{\prime }vw+u{v}^{\prime }w+uv{w}^{\prime }-u^{″}{v}^{″}{w}^{″})u^{\prime }-(u^{″}vw+u{v}^{″}w+uv{w}^{″}-u^{\prime }{v}^{\prime }{w}^{\prime })u^{″}=0$$

that can be written in the form

$$u^{\prime }w(u^{\prime }v-u{v}^{\prime })+u^{\prime }{w}^{\prime }(u^{″}{v}^{\prime }-uv)-u^{″}w(u{v}^{″}-u^{″}v)+u^{″}{w}^{″}(u^{\prime }{v}^{″}-uv)=0$$

and it is valid because of (7) and (8) and the equalities

$$\begin{array}{ccc}\hfill u^{″}{v}^{\prime }-uv& =& -(a-d)(b-c)w,\hfill \end{array}$$

(11)

$$\begin{array}{ccc}\hfill u^{\prime }{v}^{″}-uv& =& (a-d)(b-c)w.\hfill \end{array}$$

(12)

Theorem 7. 

The circumscribed circle of the diagonal triangle $UVW$ of the quadrangle $ABCD$ is incident to its center O.

The same result can be found in [2,3,6,8,10].

The line $\mathcal{U}$ has the equation $u^{″}x+u^{\prime }y=2u$ and the normal from O to this line is given by $u^{\prime }x-u^{″}y=0$. The intersection point of these two lines is the point

$$\left(\frac{2u{u}^{″}}{{u^{\prime }}^2+{u^{\prime \prime }}^2},\frac{2u{u}^{\prime }}{{u^{\prime }}^2+{u^{\prime \prime }}^2}\right).$$

(13)

Out of the equalities (8) and (12), and (9) and (11), the next equalities follow

$$(u{v}^{″}-u^{″}v)w=(u^{\prime }{v}^{″}-uv)w^{″},(u{v}^{\prime }-u^{\prime }v)w=(u^{″}{v}^{\prime }-uv)w^{\prime }$$

that can be written in the form

$$u{v}^{″}w+uv{w}^{″}-u^{\prime }{v}^{″}{w}^{″}=u^{″}vw,u{v}^{\prime }w+uv{w}^{\prime }-u^{″}{v}^{\prime }{w}^{\prime }=u^{\prime }vw.$$

(14)

The expression

$$(u^{″}vw+u{v}^{″}w+uv{w}^{″}-u^{\prime }{v}^{\prime }{w}^{\prime })u^{″}+(u^{\prime }vw+u{v}^{\prime }w+uv{w}^{\prime }-u^{″}{v}^{″}{w}^{″})u^{\prime }$$

can be written as

$$vw({u^{\prime }}^2+{u^{″}}^2)+u^{″}(u{v}^{″}w+uv{w}^{″}-u^{\prime }{v}^{″}{w}^{″})+u^{\prime }(u{v}^{\prime }w+uv{w}^{\prime }-u^{″}{v}^{\prime }{w}^{\prime }),$$

and because of (14), which is equal to $vw({u^{\prime }}^2+{u^{″}}^2)+vw{u^{″}}^2+vw{u^{\prime }}^2=2vw({u^{\prime }}^2+{u^{″}}^2)$. This means that line with the equation

$${\mathcal{W}}_o\dots (u^{″}vw+u{v}^{″}w+uv{w}^{″}-u^{\prime }{v}^{\prime }{w}^{\prime })x+(u^{\prime }vw+u{v}^{\prime }w+uv{w}^{\prime }-u^{″}{v}^{″}{w}^{″})y=4uvw$$

(15)

is incident to the point (13), the pedal of the normal to the line $\mathcal{U}$ from the point O. Because of the symmetry, it is incident to the pedals of the normal to the line $\mathcal{V}$ and $\mathcal{W}$ from the point O, respectively. Hence, the line ${\mathcal{W}}_o$ in (15) is the Wallace’s line of the point O with respect to the triangle $UVW$. Therefore, we proved our original statement, as can be seen in Figure 1:

Figure 1. The Wallace’s line ${\mathcal{W}}_O$ of the center O with respect to the triangle $UVW$ and the line $O{O}_{UVW}$ form equal angles with the asymptotes of $\mathcal{H}$.

Theorem 8. 

The Wallace’s line of the center O with respect to the diagonal triangle $UVW$ and the connecting line of the points $O_{UVW}$ and O form equal angles with the asymptotes $\mathcal{X}$ and $\mathcal{Y}$ of the hyperbola $\mathcal{H}$.

Namely, their slopes are opposite.

The line through the midpoint $(\frac{a+b}{2},\frac{a+b}{2ab})$ of the side $AB$ and parallel to the line $CD$ has the equation $x+cdy-\frac{a+b}{2ab}(ab+cd)=0$ and it is incident to the point

$$U_o=\left((ab+cd)\frac{u^{″}}{2u},(ab+cd)\frac{u^{\prime }}{2pu}\right)$$

because

$$ab{u}^{″}+u^{\prime }-(a+b)u=0.$$

(16)

Because of the symmetry of the coordinates of this point on pairs $a,b$ and $c,d$, it follows that the line incident to the midpoint of $CD$ and parallel to the side $AB$ is also incident to $U_o$. The midpoint of $AB$ and the point $U_o$ are lying on the circle given by

$$2pu(x^2+y^2)+[p(u^{\prime }-su)+u^{\prime }]x+[p^2{u}^{″}+c^2{d}^2(c+d)-a^2{b}^2(a+b)]y=0.$$

This circle is incident to the midpoint of $CD$ and obviously to the point O. There are two more such circles obtained in an analogous way. As it is stated in [1,3], the following is valid:

Theorem 9. 

The circles incident to the midpoints of $AB$, $CD$, and the point $U_o$; $AC$, $BD$, and $V_0$; $AD$, $BC$, and $W_0$ are incident to O.

The triangles $BCD$ and $ACD$ have centroids $G_a=\left(\frac{1}{3}(b+c+d),\frac{1}{3}(\frac{1}{b}+\frac{1}{c}+\frac{1}{d})\right)$, $G_b=\left(\frac{1}{3}(a+c+d),\frac{1}{3}(\frac{1}{a}+\frac{1}{c}+\frac{1}{d})\right)$ and their connecting line $G_a{G}_b$ has the equation $3cdx+3py=cds+ab(c+d)$. Analogously, the line $G_c{G}_d$ has the equation $3abx+3py=abs+cd(a+b)$. The intersection point $U_g=G_a{G}_b\cap G_c{G}_d$ is of the form

$$U_g=\left(\frac{s}{3}-\frac{u^{\prime }}{3u},\frac{c+d}{3cd}+\frac{u^{\prime }}{3abu}\right).$$

The orthocenters $H_a$ and $H_b$ from (6) have a connecting line $H_a{H}_b$ with the equation $cdpx+y=-cd(a+b)$, and analogously, the line $H_c{H}_d$ has the equation $abpx+y=-ab(c+d)$. The intersection point $U_h=H_a{H}_b\cap H_c{H}_d$ is

$$U_h=\left(-\frac{u^{\prime }}{pu},-\frac{p{u}^{″}}{u}\right).$$

Let us remember from [14] that the circumcenter of the triangle $ABC$ is the point

$$O_d=\left(\frac{1}{2}\left(a+b+c+\frac{1}{abc}\right),\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+abc\right)\right).$$

(17)

The circumcenters $O_a$ and $O_b$ with forms analogous to (17) have the connecting line $O_a{O}_b$ with the equation $cdx-y=\frac{1}{2cd}(c+d)(c^2{d}^2+1)$, and analogously, the line $O_c{O}_d$ has the equation $abx-y=\frac{1}{2ab}(a+b)(a^2{b}^2+1)$. For the intersection point $U_o=O_a{O}_b\cap O_c{O}_d$, we obtain the form

$$U_o=\left(\frac{1}{2pu}(psu-p{u}^{\prime }+u^{\prime }),\frac{1}{2pu}[ab(c+d)u+cd{u}^{\prime }+p^2{u}^{″}]\right).$$

Out of the terms for $U_g,U_h$, and $U_o$, it is easy to check that the equality $U_h+2U_o=3U_g$ is valid, i.e., $U_h-U_g=2(U_g-U_o)$ or $U_g{U}_h=2U_o{U}_g$, i.e., $U_o{U}_g:U_g{U}_h=1:2$. The same is valid for the analogous intersections. So, we have proved the result that can be found in [9], where Myakishev addressed it to J. Ganin:

Theorem 10. 

If $G_a,G_b,G_c,G_d$ are centroids, $H_a,H_b,H_c,H_d$ are orthocenters and $O_a,O_b,O_c,O_d$ are the circumcenters of the triangles $BCD,ACD,ABD,ABC$ in the quadrangle $ABCD$, and if $U_g,V_g,W_g$; $U_h,V_h,W_h$ and $U_o,V_o,W_o$ represent the diagonal points of the quadrangles $G_a{G}_b{G}_c{G}_d$, $H_a{H}_b{H}_c{H}_d$, and $O_a{O}_b{O}_c{O}_d$, respectively, then the triples of points $U_g,U_h,U_o$; $V_g,V_h,V_o$; $W_g,W_h,W_o$ are collinear and $U_o{U}_g:U_g{U}_h=V_o{V}_g:V_g{V}_h=W_o{W}_g:W_g{W}_h=1:2$ is valid.

3.3. Isogonality with Respect to the Triangles $BCD$, $ACD$, $ABD$, $ABC$

If two lines $\mathcal{L}$ and ${\mathcal{L}}^{\prime }$ have slopes $\frac{m}{n}$ and $\frac{m^{\prime }}{n^{\prime }}$, then for the oriented angle $\angle (\mathcal{L},{\mathcal{L}}^{\prime })$, the following formula is valid

$$tg\angle (\mathcal{L},{\mathcal{L}}^{\prime })=\frac{m^{\prime }n-m{n}^{\prime }}{m{m}^{\prime }+n{n}^{\prime }}.$$

(18)

The lines $AB,AC,AD$ have slopes $-\frac{1}{ab},-\frac{1}{ac},-\frac{1}{ad}.$ Let $D^{\prime }$ be the point that is isogonal to the point D with respect to the triangle $ABC$ and let k be the slope of $A{D}^{\prime }$. Then, $\angle (AB,AD)=\angle (A{D}^{\prime },AC)$ and due to (18), we obtain

$$tg\angle (AB,AD)=\frac{ad-ab}{a^{2}bd+1},tg\angle (A{D}^{\prime },AC)=\frac{ack+1}{ac-k}.$$

Out of the equality $\frac{ad-ab}{a^{2}bd+1}=\frac{ack+1}{ac-k}$, it follows that

$$k=\frac{a^{2}bc-a^{2}bd-a^{2}cd-1}{a^{3}bcd+ab+ac-ad}.$$

(19)

We will show that the point

$$D^{\prime }=\left(\frac{d-a-b-c}{abcd-1},\frac{abd+acd+bcd-abc}{abcd-1}\right)$$

is the isogonal point to the point D with respect to the triangle $ABC$. Because of the symmetry on $a,b,c$, it is enough to show that the line $A{D}^{\prime }$ is isogonal to the line $AD$ with respect to the lines $AB$ and $AC$, i.e., that the line $A{D}^{\prime }$ have the slope k from (19). The points A and $D^{\prime }$ have the difference between the coordinates

$$\begin{array}{ccc}\hfill a-\frac{d-a-b-c}{abcd-1}& =& \frac{1}{abcd-1}(a^{2}bcd+b+c-d),\hfill \\ \hfill \frac{1}{a}-\frac{abd+acd+bcd-abc}{abcd-1}& =& \frac{1}{a(abcd-1)(a^{2}bc-a^{2}bd-a^{2}cd-1)},\hfill \end{array}$$

so the line $A{D}^{\prime }$ has the slope k in (19). The point $D^{\prime }$ can be rewritten as

$$D^{\prime }=\left(\frac{2d-s}{p-1},\frac{r-2abc}{p-1}\right).$$

(20)

In the same way, we can obtain the points $A^{\prime },B^{\prime },C^{\prime }$ isogonal to the points $A,B,C$ with respect to the triangles $BCD,ACD,ABD$, respectively. The centroid of these four points is the point

$$G^{\prime }=\left(-\frac{s}{2(p-1)},\frac{r}{2(p-1)}\right).$$

(21)

The point $D^{\prime }$ from (20) and its analogous point $C^{\prime }$ have the midpoint $(-\frac{a+b}{p-1},\frac{a+b}{p-1}cd)$ that lies on the line $AB$ from (3). The line $C^{\prime }{D}^{\prime }$ has the slope $ab$; hence, it is perpendicular to the line $AB$. This means that $AB$ is the bisector of the line segment $C^{\prime }{D}^{\prime }$. Similarly, the same is valid for the analogous elements of the quadrangles $ABCD$ and $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$. Because of this, the sides $AB,AC,AD,BC,BD,CD$ of the quadrangle $ABCD$ are bisectors of the sides $C^{\prime }{D}^{\prime },B^{\prime }{D}^{\prime },B^{\prime }{C}^{\prime },A^{\prime }{D}^{\prime },A^{\prime }{C}^{\prime },A^{\prime }{B}^{\prime }$, respectively. Out of the earlier facts, it follows that the points $A,B,C,D$ are the centers of the circles $B^{\prime }{C}^{\prime }{D}^{\prime }$, $A^{\prime }{C}^{\prime }{D}^{\prime }$, $A^{\prime }{B}^{\prime }{D}^{\prime }$, $A^{\prime }{B}^{\prime }{C}^{\prime }$ that can be directly proved analytically, because for the distance of the point $D^{\prime }$ from the point $A=(a,\frac{1}{a})$, we obtain

$$a^2{(p-1)}^{2}A{D^{\prime }}^2=a^2{(a^{2}bcd+b+c-d)}^2+{[a^2(nc-bd-cd)-1]}^2=(a^2{b}^2+1)(a^2{c}^2+1)(a^2{d}^2+1),$$

so by analogy, we conclude that $A{D}^{\prime }=A{C}^{\prime }=A{B}^{\prime }$. We proved the statement found in [2,12]:

Theorem 11. 

The points $A,B,C,D$ are the centers of the circles $B^{\prime }{C}^{\prime }{D}^{\prime }$, $A^{\prime }{C}^{\prime }{D}^{\prime }$, $A^{\prime }{B}^{\prime }{D}^{\prime }$, $A^{\prime }{B}^{\prime }{C}^{\prime }$.

This is in addition to the statement found in [5]:

Theorem 12. 

The points $A^{\prime },B^{\prime },C^{\prime },D^{\prime }$ are isogonal to the points $A,B,C,D$ with respect to the triangles $BCD,ACD,ABD,ABC$ if and only if the points $A,B,C,D$ are the centers of the circles $B^{\prime }{C}^{\prime }{D}^{\prime },A^{\prime }{C}^{\prime }{D}^{\prime },A^{\prime }{B}^{\prime }{D}^{\prime },A^{\prime }{B}^{\prime }{C}^{\prime }$.

This means that the role of the quadrangle $ABCD$ for the quadrangle $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$ is the same as the role of the quadrangle $O_a{O}_b{O}_c{O}_d$ for the quadrangle $ABCD$. However, the points $A^{\prime },B^{\prime },C^{\prime },D^{\prime }$ are isogonal to the points $A,B,C,D$ with respect to the triangles $BCD,ACD,ABD,ABC$. Therefore, the following theorem is proved:

Theorem 13. 

The points $A,B,C,D$ are isogonal to the points $O_a,O_b,O_c,O_d$ with respect to the triangles $O_b{O}_c{O}_d$, $O_a{O}_c{O}_d$, $O_a{O}_b{O}_d$, $O_a{O}_b{O}_c$.

It is also stated in [2,4].

For the point $O_d$ from (17), the following equalities are valid

$$\begin{array}{ccc}\hfill x-\frac{s}{2}& =& \frac{1}{2}(\frac{1}{abc}-d)=-\frac{p-1}{2abc}\hfill \\ \hfill y-\frac{r}{2p}& =& y-\frac{1}{2}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})=\frac{1}{2}(abc-\frac{1}{d})=\frac{p-1}{2d},\hfill \\ \hfill (x-\frac{s}{2})(y-\frac{r}{2p})& =& -\frac{1}{4p}{(p-1)}^2.\hfill \end{array}$$

(22)

Hence, this point, as well as points $O_b,O_c,O_d$ are incident to the rectangular hyperbola ${\mathcal{H}}_0$ with the Equation (22) and the center $O^{\prime }=(\frac{s}{2},\frac{r}{2p})$. Due to that, $O^{\prime }$ is the center of the quadrangle $O_a{O}_b{O}_c{O}_d$ and the anticenter to $ABCD$. So, the following theorem is valid:

Theorem 14. 

The center $O^{\prime }$ of the quadrangle $O_a{O}_b{O}_c{O}_d$ is the anticenter of the quadrangle $ABCD$. The center O of this quadrangle is the anticenter of $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$.

This theorem is our original result. Its visualization is given in Figure 2.

Figure 2. The visualization of Theorems 12 and 14.

The center of the quadrangle $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$ is the point symmetric to the point O with respect to the centroid $G^{\prime }$ of this triangle, given by (21), so this center is the point $(-\frac{s}{p-1},\frac{r}{p-1})$. It is easy to see that the point $O_d$ from (17) and analogous points $O_a,O_b,O_c$ have the centroid $G_o=(\frac{s}{8p}(3p+1),\frac{r}{8p}(p+3))$. As $O^{\prime }=(\frac{s}{2},\frac{r}{2p})$ is the center of the quadrangle $O_a{O}_b{O}_c{O}_d$, the anticenter is the point symmetric to the point $O^{\prime }$ with respect to the point $G_o$ and that is the point $O_o=(\frac{s}{4p}(p+1),\frac{r}{4p}(p+1))$.

If we apply a translation for the vector $[\frac{s}{p-1},-\frac{r}{p-1}]$ on the quadrangle $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$, then, e.g., the point $D^{\prime }$ from (20) transfers to the point $D^{″}=(\frac{2d}{p-1},-\frac{2abc}{p-1})$. In the same way, we can obtain the points $A^{″},B^{″},C^{″}$. All the four points have the same product of the coordinates, so they are all incident to the rectangular hyperbola ${\mathcal{H}}^{\prime \prime }$ with the center O and with the same asymptotes as the rectangular hyperbola $\mathcal{H}$. Hence, the point O is the center of the quadrangle $A^{″}{B}^{″}{C}^{″}{D}^{″}$, so the point $(-\frac{s}{p-1},\frac{r}{p-1})$ is the center of the quadrangle $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$. The symmetric point to the latter point with respect to the centroid $G^{\prime }$ from (21) of the quadrangle $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$ is the point O.

4. Discussion

Putting the complete quadrangle into such a coordinate system that its circumscribed hyperbola is rectangular and has the equation xy = 1 allows us to prove many known properties use the same method. We use rectangular coordinates symmetrically on four vertices and four parameters $a,b,c,d$ which simplify the analytical computing. Thus, we came across some more quadrangles related to the referent one which allowed us to analyze published as well as original results.

That approach enabled us to extend our results in the rich geometry of a complete quadrangle, such as related to an isoptic point, which are planned to be presented in a future paper.