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 $(0,0)$,

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


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,


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?


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 $$\alpha - \beta + \gamma - \delta = 180^\circ \tag{7}$$

  • $\begingroup$ 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

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. $\square$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.