[alogo] Ptolemy's theorem (extension to inequality)

In a convex quadrilateral ABCD, the inequality |BC||AD|+|CD||AB| >= |BD||AC| is valid. The equality is valid if and only if the quadrilateral has a circumcircle d (is cyclic).

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The proof is based on an inversion, defined by a circle c, centered at B, and with arbitrary radius r. Such an inversion maps the circle passing through B, C and A onto a line e. In general the lengths of segments |CD| and |C'D'|, where C', D' are the inverses of C, D by an inversion with center B and radius r, is given by the formula:

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Then start from the triangle inequality |A'C'| <= |A'D'| + |D'C'| and apply the previous formula to express the lengths by means of the original side-lengths of the quadrilateral.

[alogo] Application to equilateral triangles

For an equilateral triangle ABC and a point P, the distances from the vertices satisfy PA+PB >= PC and equality occurs exactly when P is on the arc AB of the circumcircle.

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In fact, by Ptolemy, measuring quadrangle APBC: PA*BC + PB*AC >= PC*AB, which, because of the equal sides, simplifies to PA + PB >= PC. To examine the equality, consider P on the arc AB as in the figure below.

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Take P' such that PAP' is equilateral. Then triangles APB and AP'C are equal and consequently PA+PB=PP'+P'C.

See Also

Inversion.html
InverseLengths.html

References


Pedoe, D. A course of Geometry. New York, Dover, 1990, p. 111.

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