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Suppose a plane quadrilateral ABCD (convex, concave or crossed) no side of which is parallel to y-axis, and let $m_1, m_2, m_3, m_4$ be the slopes of the equations of sides AB, BC, CD, DA. Having made these definitions, now we may state the following theorem:

ABCD is a cyclic quadrilateral iff $$(m_1m_3-1)(m_2+m_4)=(m_2m_4-1)(m_1+m_3)$$

I can prove this theorem using the theory of circumscribing conics, but I would appreciate if someone could show me a different proof.

Proof based on the theory of conics:

Let $L_1\equiv m_1x -y +r_1=0$, $L_2\equiv m_2x -y +r_2=0$, $L_3\equiv m_3x -y +r_3=0$, $L_4\equiv m_4x -y +r_4=0$ be the equations of lines $AB$, $BC$, $CD$, $DA$.

Then all the conics which circumscribe the quadrilateral ABCD can be given by the equation $\lambda L_1L_3+\mu L_2L_4=0$.

Therefore all the conics circumscribing the quadrilateral are given by the equation $$\lambda(m_1x -y +r_1)(m_3x -y +r_3)+\mu(m_2x -y +r_2)(m_4x -y +r_4)=0,$$ $$(m_1m_3\lambda +m_2m_4\mu)x^2-((m_1+m_3)\lambda+(m_2+m_4)\mu)xy+(\lambda+\mu)y^2+...=0$$

If ABCD is a cyclic quadrilateral, then there is a circle circumscribing it, represented by an equation whose coefficients of $x^2$ and $y^2$ are equal and whose coefficient of $xy$ vanishes. Therefore

$$\begin {cases} (m_1m_3-1)\lambda +(m_2m_4-1)\mu=0\\ (m_1+m_3)\lambda +(m_2+m_4)\mu=0 \\ \end {cases} $$

As this system has to have a solution distinct from the trivial one $(0,0)$,

$$\begin{vmatrix} (m_1m_3-1) & (m_2m_4-1)\\ (m_1+m_3) & (m_2+m_4))\\\end{vmatrix}=0, $$ $$(m_1m_3-1)(m_2+m_4)=(m_2m_4-1)(m_1+m_3),$$

QED.

Conversely, if $$(m_1m_3-1)(m_2+m_4)=(m_2m_4-1)(m_1+m_3),$$ then the system above has a solution $(\lambda,\mu)$ distinct from the trivial one $(0,0)$. Therefore there is an ordered pair $(\lambda,\mu)\neq (0,0)$ which renders the following equation of a conic circumscribing the quadrilateral ABCD:

$$(\lambda +\mu)x^2 +0.xy+(\lambda +\mu)y^2+...=0$$

As $(\lambda +\mu)^2>0$, the conic given by this equation must be an ellipse (and a real one and non degenerate, because this conic passes through four real points), and, more precisely, a circle, because of the equal coefficients of $x^2$ and $y^2$, hence ABCD is a cyclic quadrilateral,

QED.

Note: $(\lambda +\mu)^2\neq0$ for two reasons (one algebric, the other geometric). First, because if $(\lambda +\mu)^2=0$, then $(\lambda +\mu)=0$, then $\lambda=-\mu$, then $m_1m_3=m_2m_4$ and $m_1+m_3=m_2+m_4$, then $m_1=m_2$ and $m_3=m_4$ (absurd!), or $m_1=m_4$ and $m_3=m_2$ (absurd!). Second, because if $(\lambda +\mu)^2=0$, then $(\lambda +\mu)=0$, then the equation wouldn´t have second degree terms, degrading to an equation of a straight line passing through four non collinear points (absurd!)

Is anyone acquainted with another proof?

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As always, ignoring cases where denominators may vanish ...

We may assume the quadrilateral is inscribed in the origin-centered circle of radius $k$. We can coordinatize thusly: $$\begin{array}{c} A = k (\cos 2\alpha,\sin 2\alpha) \quad B = k(\cos2\beta,\sin2\beta) \\[4pt] C=k(\cos2\gamma,\sin2\gamma) \quad D=k(\cos2\delta,\sin2\delta) \end{array} \tag{1}$$ Then we can compute successive slopes as follows: $$\begin{array}{c} \displaystyle p = \frac{k \sin 2\alpha - k \sin 2\beta}{k\cos 2\alpha-k\cos 2\beta}=\frac{\phantom{-}2\sin(\alpha-\beta)\cos(\alpha+\beta)}{-2\sin(\alpha-\beta)\sin(\alpha+\beta)}=\cot(\alpha+\beta)=\frac{1-ab}{a+b} \\[4pt] \displaystyle q=\frac{1-bc}{b+c}\qquad r=\frac{1-cd}{c+d} \qquad s=\frac{1-da}{d+a} \end{array}\tag{2}$$ where $a:=\tan\alpha$, $b:=\tan\beta$, $c:=\tan\gamma$, $d:=\tan\delta$. Eliminating $a$, $b$, $c$, $d$ from the system is straightforward, though a bit tedious, giving the result, which can be written as follows:

$$p - q + r - s = p q r s \left(\frac1p -\frac1q+\frac1r-\frac1s \right) \tag{3}$$


Note: Let's set aside the coordinates in $(1)$ and repurpose $\alpha$, etc, to write the successive slopes as $$p = \tan\alpha \qquad q= \tan\beta \qquad r = \tan\gamma \qquad s = \tan\delta \tag{4}$$ Then $(3)$ reduces to this trigonometric form $$\sin(\alpha-\beta+\gamma-\delta) = 0 \tag{5}$$ That is, for some integer $n$, $$\alpha - \beta + \gamma - \delta = 180^\circ\,n \tag{6}$$ We can see this with a bit of angle-chasing in a typical configuration:

enter image description here

Since opposite angles $180^\circ-\alpha+\delta$ and $180^\circ+\beta-\gamma$ are supplementary in a cyclic quadrilateral, we have $$\alpha - \beta + \gamma - \delta = 180^\circ \tag{7}$$

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  • $\begingroup$ And what about the proof of the reciprocal statement? The theorem has an iff, not just an if... $\endgroup$ – MrDudulex Oct 20 '18 at 13:23
  • $\begingroup$ @MrDudulex: True enough. I should've mentioned that I was only proving one way here. That said, if there's a way to pin the $n$ in (6) to $1$, then (7) essentially forms an if-and-only-if bridge (as with the other recent question, I'm ignoring complications about how the angle chase might go with more complicated configurations). $\endgroup$ – Blue Oct 20 '18 at 13:40
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Here's an alternative, but somewhat unmotivated, take.

We may assume the (extended) diagonals meet at the origin. Suppose $\overline{AC}$ and $\overline{BD}$ make respective angles of $\theta$ and $\phi$ with the $x$-axis. For some $a$, $b$, $c$, $d$, we can write $$\begin{array}{c} A = a (\cos\theta, \sin\theta)\qquad B =b (\cos\phi, \sin\phi) \\[4pt] C = c(\cos\theta, \sin\theta) \qquad D= d(\cos\phi, \sin\phi) \end{array} \tag{1}$$ Then we can calculate successive slopes $p$, $q$, $r$, $s$ as $$\begin{array}{c} \displaystyle p = \frac{a \sin\phi - b \sin\theta}{a \cos\phi - b \cos\theta} \qquad q = \frac{b \sin\theta - c \sin\phi}{b \cos\theta - c \cos\phi} \\[4pt] \displaystyle r = \frac{c \sin\phi - d \sin\theta}{c \cos\phi - d \cos\theta} \qquad s = \frac{d\sin\theta - a \sin\phi}{d\cos\theta - a \cos\phi}\end{array} \tag{2}$$ Then we have by direct substitution and simplification $$(pr-1)(q+s)-(qs-1)(p+r) = \frac{(a c - b d) (a - c) (b - d) \sin(\phi-\theta)}{(\cdots)}\tag{3}$$ where the denominator is just the product of the denominators of the slopes. We may assume that none of them are zero; that is, none of the quadrilaterals sides are "vertical".

Note that $a=c$ implies that $A$ and $C$ coincide; likewise, $b=d$ implies $B$ and $D$ coincide. Moreover, $\sin(\phi-\theta)$ implies that $A$, $B$, $C$, $D$ are collinear. Consequently,

For non-degenerate quadrilaterals with non-vertical sides, $$(pr-1)(q+s)=(qs-1)(p+r)\quad\iff\quad a c = b d \quad\iff\quad \square ABCD \text{ is cyclic}\tag{$\star$}$$

The second double-implication is guaranteed by the power of a point theorems: products $ac$ and $bd$ are equal if and only if they each compute the power of the origin with respect to some circle. $\square$

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  • $\begingroup$ Amazing proof, an almost complete one, Blue! The only type of quadrilateral not covered by your proof are the crossed quadrilaterals whose diagonals are parallel $\endgroup$ – MrDudulex Oct 20 '18 at 22:41
  • $\begingroup$ @MrDudulex: Consider that an "exercise for the reader". :) $\endgroup$ – Blue Oct 20 '18 at 22:45

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