Is the converse of Ptolemy's theorem always true I.e.

In a quadrilateral, if the product of its diagonals is equal to the sum of the products of the pairs of the opposite sides, then the quadrilateral is cyclic

if yes ; please write a proof

If no ; what are the conditions to be true

Thank you for your help


Ptolemy's inequality is just the triangle inequality under circle inversion.

If $A,B,C,D$ (in this order) are the vertices of a convex quadrilateral, then $$ AC\cdot BD \leq AB\cdot CD + BC\cdot AD \tag{1}$$ holds. If we consider a unit circle centered at $A$ and the inversion with respect to such circle, bringing $B\mapsto B'$, $C\mapsto C'$, $D\mapsto D'$, then $(1)$ is equivalent to $$ B'D' \leq B'C'+C'D'\tag{2} $$ with equality holding only if $B',C',D'$ are collinear. It follows that equality in $(1)$ holds only for cyclic quadrilaterals.


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