Ptolemy's theorem gives a relationship between the side lengths and the diagonals of a cyclic quadrilateral; it is the equality case of Ptolemy's Inequality. Proof 1 · Proof 2 (inversion) · Problems

The inequality states that in for four points $A, B, C, D$ in the plane, $AB \cdot CD + BC \cdot DA \ge AC \cdot BD$ , with equality if and only if $ABCD$ is a

Multiple Proofs? I think the proof where you draw the diagonals and flip one of the triangles to get a quadrilateral ABCD' and then use the (1/2)ab sin C ...

Solution 1 (Ptolemy's Theorem). Ptolemy's theorem states that for cyclic quadrilateral $WXYZ$ , $WX\cdot YZ + XY\cdot WZ = WY\cdot XZ$ . We may assume that ...

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Solution 6 (Ptolemy's Theorem). [asy] pathpen = black; pointpen = black; size(6cm);. Let $s = 200$ . Let $O$ be the center of the circle. Then $AC$ is twice ... Solution 2 (Algebra) · Solution 6 (Ptolemy's Theorem)

Solution. Since quadrilateral $ABMC$ is inscribed in circle $O$ , thus it is a cyclic quadrilateral. By Ptolemy's Theorem, \[AC \cdot MB + MC \cdot AB = BC ...

4 февр. 2017 г. · Ptolemy's theorem is a relation between the sides and diagonals of a cyclic quadrilateral. In this article, we go over the uses of the theorem ...

We can use the Pythagorean Theorem and similar triangles to find Points $A$ , $B$ , $D$ , and $F$ all lie on a circle whose diameter is $AB$.

A cyclic quadrilateral is a quadrilateral that can be inscribed in a circle. While all triangles are cyclic, the same is not true of quadrilaterals.