Let $ABCD$ be a quadrilateral in the standard Euclidean plane, with side lengths $\overline{\rm AB}=a$, $\overline{\rm BC}=b$, $\overline{\rm CD}=c$, $\overline{\rm DA}=d$, and diagonals of length $\overline{\rm AC}=m$ and $\overline{\rm BD}=n$. Then it is a well known theorem of elementary plane geometry that if $ABCD$ is a cyclic, the following relation holds (see Ptolemy's Theorem for a proof):
\begin{equation}
\frac{m}{n}= \frac{ad+bc}{ab+cd} \quad (I).
\end{equation}
This statement is the so called "Ptolemy's Second Theorem", even though it is named after Legendre in Italian textbooks on geometry for some obscure reason. Now, as in the case of Ptolemy's (First) Theorem, I would like to show that this theorem has a converse, too.
More precisely, I would like to show that if a quadrilateral ABCD is such that (I) holds, then it must be cyclical. I could not find a proof, but I am pretty sure that it is true. Any help is welcome.
NOTE. The following intuitive argument shows that the converse of Ptolemy's Second Theorem is for sure true under the additional hypothesis that $ABCD$ is convex. To see why, assume that $ABCD$ is not cyclical, so that e.g. $\angle BAD + \angle BCD < \pi$, and $a+d \geq b+c$ (if instead $a+d < b+c$ switch the roles of $AB$ and $CD$ in what follows). Then we can deform the original quadrilateral $ABCD$ by rotating the point $D$ around $A$ and the point $C$ around $B$ so to increase the angle $\angle BAD$, to preserve the side lengths and to make the new quadrilateral $ABC'D'$ cyclic (to see that this is possible, note that, since $a+d \geq b+c$, we can deform $ABCD$ by moving $C$ and $D$ so that the points $B$, $C''$ and $D''$ are aligned, in which case the sum of the angles $\angle BAD''$ and $\angle BC''D''$ is greater than $\pi$, so by continuity there are $C'$ and $D'$ such that $\angle BAD' + \angle BC'D' = \pi$). But then it is easy to check that $\overline{\rm AC'} < m$ and $\overline{\rm BD'} > n$, so that by applying Legendre's Theorem to $ABC'D'$ we conclude that (I) cannot hold. But I am searching for a general proof that covers also the case in which $ABCD$ is concave.
Best Answer
Let $x\approx 4.45171859943238$ be the root of the function $$f(t)=\frac{2}{t-\sqrt 3} - \frac{4\sqrt{t^2+1}}{t^2+5}.$$ Let $ABC$ be an equilateral triangle with sides of length $2$. Let $D$ be a point on the same side of line $AC$ as $B$ such that $AD=CD=\sqrt{x^2+1}$. Then the distance of $D$ from $AC$ is $x$, hence $BD=x-\sqrt 3$. We have $$\frac{AC}{BD} = \frac{2}{x-\sqrt 3} = \frac{4\sqrt{x^2+1}}{x^2+5} = \frac{AB \cdot AD + CB \cdot CD}{BA \cdot BC + DA \cdot DC},$$ meaning that $ABCD$ is a counterexample to the statement you want to prove.