Isoptic Point of the Non-Cyclic Quadrangle in the Isotropic Plane

Abstract

We study the non-cyclic quadrangle $ABCD$ in the isotropic plane and its isoptic point. This is a continuation of the research carried out in a few previous papers. There, we put the non-cyclic quadrangle in the standard position, which enables us to prove its properties using a simple analytical method. In the standard position, the special hyperbola $xy=1$ is circumscribed to the quadrangle. Hereby, we use the same method to obtain several results related to the isoptic point of the non-cyclic quadrangle. The isoptic point T is the inverse image of points $A^{\prime }$, $B^{\prime }$, $C^{\prime }$, $D^{\prime }$ with respect to circumcircles of $BCD$, $ACD$, $ABD$, $ACD$, respectively, where $A^{\prime }$, $B^{\prime }$, $C^{\prime }$, $D^{\prime }$ are isogonal points to vertices A, B, C, D with respect to triangles $BCD$, $ACD$, $ABD$, $ACD$. The circumircles are seen from T under the equal angles. Our analysis is motivated by the Euclidean results already published in the literature.

1. Introduction

If four points are joined in pairs by six distinct lines, we study the figure called a complete quadrangle [1,2,3]. There, we chose an affine coordinate system in such a way that enabled us to put every non-cyclic quadrangle in the standard position. Hence, in order to prove a geometric property for any quadrangle, it is sufficient to prove that the fact holds for a standard quadrangle. Using this simple analytical method, we can expand furthermore the investigation on a non-cyclic quadrangle. Here, we focus on the properties related to the isoptic point of the quadrangle. The motivation for this study is found in the Euclidean case.

In [4], we studied the geometry of the complete quadrangle in the Euclidean plane related to its isoptic point. We proved well-known facts found in the available literature, such as [5,6,7,8], and extended those to obtain original results. In that study, we used the same analytical method for all the proofs. We have showed that the isoptic point of the complete quadrangle is the inverse image of isogonal points to vertices of the quadrangle with respect to four triangles related to the quadrangle. That point has the property that circumcircles of triangles are seen under the equal angles. This property is visualized in Figure 1 from [4]. In this paper, we extend these properties further by studying them in the isotropic plane.

2. Materials and Methods

Let us recall that the isotropic plane is a real projective metric plane. The metric is induced by the absolute figure $(\mathsf{\Omega },\omega )$, the point $\mathsf{\Omega }$ representing the absolute point, and the line $\omega$ representing the absolute line. When studying the projective plane in homogeneous coordinates where any point is in the form $T=(x_0:x_1:x_2)$, usually, a projective coordinate system is chosen such that $\mathsf{\Omega }=(0:1:0)$ and $\omega$ with equation $x_2=0$.

Now, we list important facts regarding the isotropic plane [9]. Points incident to $\omega$ are called the isotropic points and lines incident to $\mathsf{\Omega }$ are isotropic lines. Parallel lines have the same isotropic point and parallel points are incident to one isotropic line.

Further, in accordance with the projective coordinate system, in an affine model of the isotropic plane, we take $x={\displaystyle \frac{x_0}{x_2}}$ and $y={\displaystyle \frac{x_1}{x_2}}$. For two non-parallel points $A=(x_A,y_A)$ and $B=(x_B,y_B)$, a distance is defined by $d(A,B):=x_B-x_A$. The span $s(A,B):=y_B-y_A$ is defined for parallel points $A=(x,y_A)$ and $B=(x,y_B)$. Both of these quantities are signed. The angle formed by two lines $p_1,p_2$ in the form $y=k_{i}x+l_i$, $i=1,2$, is $\angle (p_1,p_2):=k_2-k_1$, being signed as well. Two points $A=(x_A,y_A)$ and $B=(x_B,y_B)$ have the midpoint $M=\left({\displaystyle \frac{x_A+x_B}{2}},{\displaystyle \frac{y_A+y_B}{2}}\right)$ and two lines $p_i$ with the equation $y=k_{i}x+l_i$, $i=1,2$, have the bisector $y={\displaystyle \frac{1}{2}}\left(k_1+k_2\right)x+{\displaystyle \frac{1}{2}}\left(l_1+l_2\right)$.

The complete classification of conics in the isotropic plane can be seen in [9]. The conic that touches the absolute line $\omega$ at the point $\mathsf{\Omega }$ is called a circle, and it is given in the affine model of the isotropic plane by $y=u{x}^2+vx+w$, $u\ne 0$, $u,v,w\in \mathbb{R}$. Its radius is defined as $\rho ={\displaystyle \frac{1}{2u}}$. If the mentioned conic intersects the line $\omega$ at the absolute point $\mathsf{\Omega }$ and in any other point, then the conic is a special hyperbola. In such a case, one asymptote is isotropic while the other is a non-isotropic line.

In [1], without loss of generality, we chose the affine coordinate system such that the center of the special hyperbola coincides with the origin of the coordinate system, and asymptotes are the x-axis and y-axis. We showed as well that such a special hyperbola can be shown in the form $xy=1$. We also proved the following important theorem that enables the study of the geometry of non-cyclic quadrangles using a simple analytical method.

Theorem 1

([1], p. 2). Any non-cyclic quadrangle $ABCD$, by the appropriate choice of an affine coordinate system, has the vertices

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

(1)

and the sides

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

(2)

and the circumscribed special hyperbola

$$\mathcal{H}\dots xy=1.$$

(3)

The following symmetric functions are very useful:

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

(4)

The Euler center

$$O=\left(0,0\right)$$

(5)

is the intersection point of Euler circles of triangles $BCD$, $ACD$, $ABD$, $ABC$. We call it the center of the quadrangle for short.

The medial point

$$N=\left(0,{\displaystyle \frac{r}{2p}}\right)$$

(6)

is defined as the intersection point of bisectors of the pairs of opposite sides of the quadrangle (see Figure 2). The focal line

$$\mathcal{M}\dots y=0$$

(7)

is the common tangent of the inscribed circles of triangles $BCD$, $ACD$, $ABD$, $ABC$.

In further research in [3], we introduced the diagonal triangle of the non-cyclic quadrangle and its properties. Hence, the diagonal points of the non-cyclic quadrangle, depicted in Figure 2, are as follows:

$$\begin{array}{c}\hfill U=AB\cap CD=\left({\displaystyle \frac{ab(c+d)-cd(a+b)}{ab-cd}},{\displaystyle \frac{a+b-c-d}{ab-cd}}\right),\\ \hfill V=AC\cap BD=\left({\displaystyle \frac{ac(b+d)-bd(a+c)}{ac-bd}},{\displaystyle \frac{a+c-b-d}{ac-bd}}\right),\\ \hfill W=AD\cap BC=\left({\displaystyle \frac{ad(b+c)-bc(a+d)}{ad-bc}},{\displaystyle \frac{a+d-b-c}{ad-bc}}\right).\end{array}$$

(8)

In the text that follows, we expand the study of the geometry of the non-cyclic quadrangle and present the obtained results related to its isoptic point.

3. Results

The circumcircles ${\mathcal{K}}_a$, ${\mathcal{K}}_b$, ${\mathcal{K}}_c$, ${\mathcal{K}}_d$ of the triangles $BCD$, $ACD$, $ABD$, $ABC$ with the equations

$$\begin{array}{cc}\hfill {\mathcal{K}}_a\dots & \hfill bcdy=x^2-(b+c+d)x+bc+bd+cd,\\ \hfill {\mathcal{K}}_b\dots & \hfill acdy=x^2-(a+c+d)x+ac+ad+cd,\\ \hfill {\mathcal{K}}_c\dots & \hfill abdy=x^2-(a+b+d)x+ab+ad+bd,\\ \hfill {\mathcal{K}}_d\dots & \hfill abcy=x^2-(a+b+c)x+ab+ac+bc\end{array}$$

(9)

According to [9] (p. 81), we have the radii

$${\rho }_a={\displaystyle \frac{bcd}{2}}={\displaystyle \frac{p}{2a}},{\rho }_b={\displaystyle \frac{acd}{2}}={\displaystyle \frac{p}{2b}},{\rho }_c={\displaystyle \frac{abd}{2}}={\displaystyle \frac{p}{2c}},{\rho }_d={\displaystyle \frac{abc}{2}}={\displaystyle \frac{p}{2d}}.$$

(10)

Let us recall that if two circles ${\mathcal{K}}_1$ and ${\mathcal{K}}_2$ intersect in two real points $P_1$ and $P_2$, then the angle between ${\mathcal{K}}_1$ and ${\mathcal{K}}_2$ at $P_1$ is opposite the angle between ${\mathcal{K}}_1$ and ${\mathcal{K}}_2$ at $P_2$, as it is explained in [9] (pp. 33–34).

Theorem 2.

Let $ABCD$ be a non-cyclic quadrangle. The angles of intersection between two circumcircles of the triangles $BCD$, $ACD$, $ABD$, $ABC$ are equal to the angles of intersection of two other circumcircles.

Proof. 

Let us prove that the angles of intersection of ${\mathcal{K}}_a$ and ${\mathcal{K}}_b$ are equal to the angles of intersection of ${\mathcal{K}}_c$ and ${\mathcal{K}}_d$. The circles ${\mathcal{K}}_a$ and ${\mathcal{K}}_b$ intersect in the points C and D. The tangent $t_{aC}$ to ${\mathcal{K}}_a$ at C is given by the equation ${\displaystyle y=-\frac{b+d-c^2}{bd}x+\frac{bd+bc+cd-c^2}{bcd}}$, while tangent $t_{bC}$ to ${\mathcal{K}}_b$ at C is given by the equation ${\displaystyle y=-\frac{a+d-c^2}{ad}x+\frac{ad+ac+cd-c^2}{acd}}$. The angle of intersection of ${\mathcal{K}}_a$ and ${\mathcal{K}}_b$ at C equals ${\displaystyle \angle \left(t_{aC},t_{bC}\right)=-\frac{a+d-c^2}{ad}+\frac{b+d-c^2}{bd}=-\frac{(a-b)(c-d)}{p}}$. The angle of intersection of ${\mathcal{K}}_a$ and ${\mathcal{K}}_b$ at D equals ${\displaystyle \frac{(a-b)(c-d)}{p}}$. The same two angles would be obtained as the angles of intersection of ${\mathcal{K}}_c$ and ${\mathcal{K}}_d$ at A and B. □

The line isogonal to $AD$ with respect to $AB$ and $AC$ is given by $y={\displaystyle \frac{bc-(b+c)d}{abcd}}x+{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}}-{\displaystyle \frac{1}{d}}$, while the line isogonal to $CD$ with respect to $AC$ and $CB$ is given by ${\displaystyle y=\frac{ab-(a+b)d}{abcd}x+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{d}}$. They obviously intersect at the point ${\displaystyle D^{\prime }=\left(0,\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{d}\right)}$, the isogonal point of D with respect to the triangle $ABC$. Therefore, the points $A^{\prime }$, $B^{\prime }$, $C^{\prime }$, $D^{\prime }$ isogonal to the vertices A, B, C, D, with respect to the triangles $BCD$, $ACD$, $ABD$, $ABC$, respectively, are

$$\begin{array}{cc}\hfill A^{\prime }=\left(0,-{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}}+{\displaystyle \frac{1}{d}}\right),& B^{\prime }=\left(0,{\displaystyle \frac{1}{a}}-{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}}+{\displaystyle \frac{1}{d}}\right),\hfill \\ \hfill C^{\prime }=\left(0,{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}-{\displaystyle \frac{1}{c}}+{\displaystyle \frac{1}{d}}\right),& D^{\prime }=\left(0,{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}}-{\displaystyle \frac{1}{d}}\right).\hfill \end{array}$$

(11)

We notice that the four isogonal points are parallel.

The points

$$A_d=\left(a,{\displaystyle \frac{b+c-a}{bc}}\right),A_c=\left(a,{\displaystyle \frac{b+d-a}{bd}}\right),A_b=\left(a,{\displaystyle \frac{c+d-a}{cd}}\right)$$

(12)

are parallel to the point A and incident to the sides $BC$, $BD$, and $CD$ of the triangle $BCD$. They are analogous to the feet of perpendiculars from A to the sides of $BCD$ in the Euclidean plane.

Theorem 3.

Let $ABCD$ be a non-cyclic quadrangle. For the points $A_b$, $A_c$, $A_d$ parallel to the point A and incident to the sides of the triangle $BCD$, the equalities

$${\displaystyle \frac{d(A,B)d(C,D)}{s(A_c,A_d)}}={\displaystyle \frac{d(A,C)d(B,D)}{s(A_b,A_d)}}={\displaystyle \frac{d(A,D)d(B,C)}{s(A_b,A_c)}}$$

are valid. Three more analogous statements are also valid.

Proof. 

From (1) and (12), we obtain ${\displaystyle \frac{d(A,B)d(C,D)}{s(A_c,A_d)}}={\displaystyle \frac{(b-a)(d-c)}{{\displaystyle \frac{b+c-a}{bc}}-{\displaystyle \frac{b+d-a}{bd}}}}=bcd$. □

Similar to (12), the coordinates of the points

$$\begin{array}{c}\hfill B_a=\left(b,{\displaystyle \frac{c+d-b}{cd}}\right),B_c=\left(b,{\displaystyle \frac{a+d-b}{ad}}\right),B_d=\left(b,{\displaystyle \frac{a+c-b}{ac}}\right),\\ \hfill C_a=\left(c,{\displaystyle \frac{b+d-c}{bd}}\right),C_b=\left(c,{\displaystyle \frac{a+d-c}{ad}}\right),C_d=\left(c,{\displaystyle \frac{a+b-c}{ab}}\right),\\ \hfill D_a=\left(d,{\displaystyle \frac{b+c-d}{bc}}\right),D_b=\left(d,{\displaystyle \frac{a+c-d}{ac}}\right),D_c=\left(d,{\displaystyle \frac{a+b-d}{ab}}\right)\phantom{,}\end{array}$$

(13)

parallel to the points B, C, D and incident to the sides of the triangles $ACD$, $ABD$, $ABC$, respectively, are obtained.

Theorem 4.

Let $ABCD$ be a non-cyclic quadrangle and the triples of points $A_b$, $A_c$, $A_d$; $B_a$, $B_c$, $B_d$; $C_a$, $C_b$, $C_d$; and $D_a$, $D_b$, $D_c$ be parallel to the points A, B, C, D and incident to the sides of the triangles $BCD$, $ACD$, $ABD$, $ABC$, respectively. The points $A_d{C}_d\cap A_c{D}_c$, $A_d{B}_d\cap A_b{D}_b$, $A_b{C}_b\cap A_c{B}_c$ are parallel points.

Proof. 

Lines $A_d{C}_d$ and $A_c{D}_c$ have the equations ${\displaystyle y=\frac{b-a-c}{abc}x+\frac{2}{b}}$ and ${\displaystyle y=\frac{b-a-d}{abd}x+\frac{2}{b}}$, and they intersect at the point ${\displaystyle A_d{C}_d\cap A_c{D}_c=\left(0,\frac{2}{b}\right)}$. Similarly, we obtain ${\displaystyle A_d{B}_d\cap A_b{D}_b=\left(0,\frac{2}{c}\right)}$ and ${\displaystyle A_b{C}_b\cap A_c{B}_c=\left(0,\frac{2}{d}\right)}$. □

The points $A^{\prime },B^{\prime },C^{\prime },D^{\prime }$ isogonal to the vertices A, B, C, D, with respect to the triangles $BCD$, $ACD$, $ABD$, $ABC$, respectively, are depicted in Figure 3.

Theorem 5.

Let $ABCD$ be a non-cyclic quadrangle and $A^{\prime },B^{\prime },C^{\prime },D^{\prime }$ be points isogonal to the vertices A, B, C, D, with respect to the triangles $BCD$, $ACD$, $ABD$, $ABC$, respectively. The centers of the quadrangles $ABC{D}^{\prime }$, $ABD{C}^{\prime }$, $ACD{B}^{\prime }$, $BCD{A}^{\prime }$ are parallel to the points D, C, B, A, respectively.

Proof. 

The special hyperbola passing through the points A, B, C, $D^{\prime }$ has the equation $d{x}^2+abcxy-(a+b+c)dx-dy-abc+abd+acd+bcd=0.$ Its center ${\displaystyle \left(d,\frac{(a+b+c-2d)d}{abc}\right)}$ is parallel to ${\displaystyle D=\left(d,\frac{1}{d}\right)}$. □

Theorem 6.

Let $ABCD$ be a non-cyclic quadrangle and $A^{\prime },B^{\prime },C^{\prime },D^{\prime }$ be points isogonal to the vertices A, B, C, D, with respect to the triangles $BCD$, $ACD$, $ABD$, $ABC$, respectively. The midpoint of the points $A^{\prime }$ and $B^{\prime }$ lies on the lines $CD=A_b{B}_a$, $A_c{B}_d$, and $A_d{B}_c$, where $A_b,A_c,A_d$ and $B_a,B_c,B_d$ are parallel points to A and B and incident to the sides of the triangles $BCD$ and $ACD$, respectively. The five analogous statements are also valid.

Proof. 

The midpoint ${\displaystyle M_{A^{\prime }{B}^{\prime }}=\left(0,\frac{1}{c}+\frac{1}{d}\right)}$ is obviously incident to the line $CD$ given in (2). It follows from (13) that the line $A_d{B}_c$ has the equation ${\displaystyle y=\frac{cd-ad-bc}{p}x+\frac{1}{c}+\frac{1}{d}}$, and the coordinates of $M_{A^{\prime }{B}^{\prime }}$ satisfy that equation. We similarly prove that $M_{A^{\prime }{B}^{\prime }}$ lies on $A_c{B}_d$. □

In the rest of the paper, we study the so-called isoptic point of the quadrangle. The polar line of the point $D^{\prime }$ to the circle ${\mathcal{K}}_d$ has the equation ${\displaystyle y=-\frac{a+b+c}{abc}x+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}$. Thus, the inverse image of $D^{\prime }$ with respect to ${\mathcal{K}}_d$ is the point ${\displaystyle T=\left(0,\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)}$. Because of the symmetry on $a,b,c,d$, we conclude that the same point is the inverse image of $A^{\prime }$, $B^{\prime }$, $C^{\prime }$ with respect to ${\mathcal{K}}_a$, ${\mathcal{K}}_b$, ${\mathcal{K}}_c$, respectively. The point

$$T=\left(0,{\displaystyle \frac{q}{p}}\right)$$

(14)

is called the isoptic point of the quadrangle $ABCD$ (see Figure 3). The property described in Theorem 12 justifies this name.

Theorem 7.

Let $ABCD$ be a non-cyclic quadrangle. The isoptic point of the quadrangle is parallel to its center and its medial point. The medial point is the midpoint of the center and the isoptic point.

Theorem 8.

Let $ABCD$ be a non-cyclic quadrangle, the point O its center, and T its isoptic point. The lines connecting the center O and the isoptic point T with a diagonal point of the quadrangle $ABCD$ are isogonal with respect to the quadrangle’s sides intersecting in this diagonal point.

Proof. 

Let us prove that the lines $UO$ and $UT$ are isogonal with respect to the sides $AB$ and $CD$ intersecting in U. The bisector of the sides $AB$ and $CD$ is given by the equation ${\displaystyle y=-\frac{ab+cd}{2p}x+\frac{r}{2p}}$. The line $UO$ has the equation ${\displaystyle y=\frac{a+b-c-d}{ab(c+d)-cd(a+b)x}}$, while the line $UT$ has the equation ${\displaystyle y=\frac{p(a+b-c-d)-r(ab-cd)}{p\left(ab(c+d)-cd(a+b)\right)}x+\frac{r}{p}}$. The bisector of $UT$ and $UO$ is, therefore, given by the equation ${\displaystyle y=-\frac{ab+cd}{2p}x+\frac{r}{2p}}$. Thus, the pairs of lines $UO$, $UT$ and $AB$, $CD$ share the same bisector, which means that $UO$ and $UT$ are isogonal with respect to $AB$ and $CD$. □

The visualization of Theorem 8 is given in Figure 4.

Theorem 9.

Let $ABCD$ be a non-cyclic quadrangle and T its isoptic point. The following statements

$$\begin{array}{cc}& \angle ATB=\angle ACB+\angle ADB,\angle ATC=\angle ABC+\angle ADC,\\ & \angle ATD=\angle ABD+\angle ACD,\angle BTC=\angle BAC+\angle BDC,\\ & \angle BTD=\angle BAD+\angle BCD,\angle CTD=\angle CAD+\angle CBD\end{array}$$

are valid.

Proof. 

The lines $AT$, $BT$ have the equations ${\displaystyle y=-\frac{bc+bd+cd}{p}x+\frac{r}{p}}$, ${\displaystyle y=-\frac{ac+ad+cd}{p}x+\frac{r}{p}}$, respectively, and they form an angle ${\displaystyle \angle ATB=\frac{(b-a)(c+d)}{p}}$. On the other hand, ${\displaystyle \angle ACB+\angle ADB=-\frac{1}{bc}+\frac{1}{ac}-\frac{1}{bd}+\frac{1}{ad}=\frac{(b-a)(c+d)}{p}}$. Thus, $\angle ATB=\angle ACB+\angle ADB$. □

Theorem 10.

Let $ABCD$ be a non-cyclic quadrangle and T its isoptic point. The following statements

$$\begin{array}{cc}& \angle ACB:\angle ADB=d(T,D):d(T,C),\angle ABC:\angle ADC=d(T,D):d(T,B),\\ & \angle ABD:\angle ACD=d(T,C):d(T,B),\angle BAC:\angle BDC=d(T,D):d(T,A),\\ & \angle BAD:\angle BCD=d(T,C):d(T,A),\angle CAD:\angle CBD=d(T,B):d(T,A)\phantom{,}\end{array}$$

are valid.

Proof. 

From (2), we obtain ${\displaystyle \frac{\angle ACB}{\angle ADB}=\frac{-{\displaystyle \frac{1}{bc}}+{\displaystyle \frac{1}{ac}}}{-{\displaystyle \frac{1}{bd}}+{\displaystyle \frac{1}{ad}}}=\frac{d}{c}=\frac{d(T,D)}{d(T,C)}}$, and we similarly prove the other five equalities. □

Let $\overline{A}$, $\overline{B}$, $\overline{C}$, $\overline{D}$ be points on the special hyperbola $\mathcal{H}$ diametrically opposite to the points A, B, C, D, i.e.,

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

We can notice that the lines $AD$ and $A\overline{D}$ are antiparallel with respect to the asymptotes of $\mathcal{H}$. Indeed, since $AD$, $A\overline{D}$ have the equations ${\displaystyle y=-\frac{1}{ad}x+\frac{a+d}{ad}}$, ${\displaystyle y=\frac{1}{ad}x+\frac{d-a}{ad}}$, their slopes are the opposite numbers.

Theorem 11.

Let $ABCD$ be a non-cyclic quadrangle with the isoptic point T and the circumscribed hyperbola $\mathcal{H}$. Let $\overline{A},\overline{B},\overline{C},\overline{D}$ be points on the hyperbola $\mathcal{H}$ diametrically opposite to the points $A,B,C,D$, respectively. The point T is isogonal to $\overline{A},\overline{B},\overline{C},\overline{D}$ with respect to the triangles $BCD,ACD,ABD,ABC$, respectively.

Proof. 

Since ${\displaystyle \angle TAC=-\frac{1}{ac}+\frac{bc+cd+bd}{abcd}=-\frac{1}{ab}+\frac{1}{ad}=\angle BA\overline{D}}$, the lines $AT$, $A\overline{D}$ are isogonal with respect to $AC$, $AB$. Similar equalities are obtained for the lines passing through the vertices B and C. Therefore, T is isogonal point to $\overline{D}$ with respect to the triangle $ABC$. □

Theorem 12.

Let $ABCD$ be a non-cyclic quadrangle and let its isoptic point T lie in the exterior of the circles $BCD$, $ACD$, $ABD$, $ABC$. These circles are seen from T under the equal angles.

In the case when the vertices of the quadrilateral are given by (1), these angles are equal to ${\displaystyle \frac{4}{\sqrt{-abcd}}}$.

Proof. 

The polar line of the point T with respect to the circle ${\mathcal{K}}_d$ given in (9) has the equation ${\displaystyle y=-\frac{a+b+c}{abc}x+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{d}}$ and it intersects ${\mathcal{K}}_d$ at two points $T_{1,2}$ whose abscissas satisfy the equation $x^2=-{\displaystyle \frac{p}{d^2}}$. Thus, T is in the exterior of ${\mathcal{K}}_d$ precisely when $p<0$, and in that case, T is in the exterior of all four circumcircles ${\mathcal{K}}_a$, ${\mathcal{K}}_b$, ${\mathcal{K}}_c$, ${\mathcal{K}}_d$. The contact points $T_{1,2}$ have the coordinates ${\displaystyle \left(\pm \frac{\sqrt{-p}}{d},\mp \frac{(a+b+c)\sqrt{-p}}{p}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{d}\right)}$, and the tangents $t_{1,2}$ from T to ${\mathcal{K}}_d$ have the equations ${\displaystyle y=\frac{\mp (a+b+c)\sqrt{-p}-2abc}{\pm abc\sqrt{-p}}x+\frac{r}{p}}$. Therefore, $t_1$ and $t_2$ form the angle ${\displaystyle \angle (t_1,t_2)=\frac{4}{\sqrt{-p}}}$. Due to the symmetry on $a,b,c,d$, we conclude that ${\mathcal{K}}_a$, ${\mathcal{K}}_b$, ${\mathcal{K}}_c$ are seen from T under the same angles ${\displaystyle \frac{4}{\sqrt{-p}}}$. □

Theorem 13.

Let $ABCD$ be a non-cyclic quadrangle, T its isoptic point, and ${\rho }_a$, ${\rho }_b$, ${\rho }_c$, ${\rho }_d$ radii of the circumcircles $BCD$, $ACD$, $ABD$, $ABC$. Then, the following equalities are valid:

$${\rho }_a·d(T,A)={\rho }_b·d(T,B)={\rho }_c·d(T,C)={\rho }_d·d(T,D).$$

Proof. 

According to (10), we obtain ${\displaystyle {\rho }_a·d(T,A)=\frac{p}{2a}·a=\frac{p}{2}}$, and the same value is obtained for the other three vertices. □

From the proof of Theorem 12, we have that the power ${\pi }_d$ of the point T with respect to the circle ${\mathcal{K}}_d$ equals ${\displaystyle {\left(d(T,T_1)\right)}^2=-\frac{p}{d^2}}$. As a consequence, we obtain the following theorem:

Theorem 14.

Let $ABCD$ be a non-cyclic quadrangle and T its isoptic point. The product of the powers of T with respect to the circumcircles $BCD$, $ACD$, $ABD$, $ABC$ is equal to the square of the product of the distances from T to A, B, C, D, i.e.,

$${\pi }_a{\pi }_b{\pi }_c{\pi }_d={\left(d(T,A)d(T,B)d(T,C)d(T,D)\right)}^2.$$

The powers of T with respect to the circumcircles $BCD$, $ACD$, $ABD$, $ABC$ are proportional to the squares of radii of the circles, i.e.,

$${\pi }_a:{\pi }_b:{\pi }_c:{\pi }_d={\rho }_a^2:{\rho }_b^2:{\rho }_c^2:{\rho }_d^2.$$

Theorem 15.

Let $ABCD$ be a non-cyclic quadrangle, T its isoptic point, and $A^{\prime }$, $B^{\prime }$, $C^{\prime }$, $D^{\prime }$ points isogonal to A, B, C, D with respect to the triangles $BCD$, $ACD$, $ABD$, $ABC$, respectively. Then,

$$d(T,A)s(T,A^{\prime })=d(T,B)s(T,B^{\prime })=d(T,C)s(T,C^{\prime })=d(T,D)s(T,D^{\prime }).$$

Proof. 

It follows directly from (1), (11), and (14). □

4. Discussion

In this paper, we extend our investigation on the geometry of the non-cyclic quadrangle in the isotropic plane with a focus on properties related to its isoptic point. Putting the non-cyclic quadrangle in the standard position (see [1]) enables us to transfer the properties from Euclidean plane to the isotropic plane. All the properties that include the terms perpendicularity and center of a circle decay in the isotropic plane.

The Euclidean versions of Theorems 2 and 11 can be found in [7], Theorems 3–6 in [10], Theorems 8–10 and 12–14 in [5], and Theorem 15 in [6].

Studying the geometry of the complete quadrangle can be expanded on the research of its dual figure, the complete quadrilateral, both in the Euclidean plane and the isotropic one. These geometries are as rich as the geometry of a triangle. Although some properties cannot be transferred to the isotropic plane, from our experience, new results not valid in the Euclidean plane can come up. Going in this direction, for now, we are working on a review paper on the properties valid for the quadrilateral in the Euclidean plane.