Ptolemy's inequality for a convex quadrilateral ABCD: AB·CD + BC·DA ≥ AC·BD with equality iff ABCD is a cyclic quadrilateral (meaning A, B, C, D lie on a circle in that order).

Proof

Take E inside ABCD such that angle DAE = angle CAB and angle ADE = angle ACB.  Then ADE and ACB are similar, so DE/CB = AD/AC and hence BC·AD = AC·DE.  It also follows that AE/AB = AD/AC.  But we also have angle EAB =  angle DAC and hence AEB and ADC are also similar. So EB/AB = DC/AC, and hence AB·CD = AC·EB. Adding, we have: AB·CD + BC·AD = AC(BE + ED) ≥ AC·BD with equality iff E lies on BD, or equivalently ABCD is cyclic.

This glosses over one point. It only follows that angle EAB = angle DAC if ABCD is convex.  For the convex case, we have that angle EAB = angle CAB + angle EAC and angle DAC = angle DAE + angle EAC, or angle EAB = angle CAB - angle EAC and angle DAC = angle DAE - angle EAC.  Either way angle EAB = angle DAC.  But in the non-convex case, we can have angle EAB = angle CAB + angle EAC and angle DAC = angle DAE - angle EAC (or - ... +) and hence the angles angle EAB and angle DAC are not necessarily equal.

Application of Ptolemy's Inequality

From Nick's Mathematical Puzzles, puzzle 139: Three Towns:

The towns of Alpha, Beta, and Gamma are equidistant from each other. If a car is three miles from Alpha and four miles from Beta, what are the maximum and minimum possible distances of the car from Gamma? Assume the land is flat.

Here I'll call the cities A, B and G for short, and the car is at C.  Let x = CG, and let y = AB = AG = BG.  When x is at its smallest possible value, it lies on the other side of AG from B (because the car is farthest from B), so the vertices of the convex quadrilateral, in order, are ABGC.  By Ptolemy's inequality,

AB·GC + BG·CA ≥ AG·BC, so xy + 3y ≥ 4y.

When the car is farthest from G, it's on the other side of AB from G, so the vertices, in order, are ACBG, giving us

AC·BG + CB·GA ≥ AB·CG, so 3y + 4y ≥ xy.

The variable, y, was kind enough to drop out of this inequality, leaving us with x + 3 ≥ 4 and 3 + 4 ≥ x, or 1 ≤ x ≤ 7.

Interestingly, if you view the car and Alpha as fixed objects three miles apart, and the cities Beta and Gamma as movable objects, then Beta lies on a circle centered at the car with radius 4.  Now consider an equilateral triangle whose vertices are the car, Alpha, and "G".  The surprise is that Gamma lies on a circle whose center is G and whose radius is 4.

Internet References

http://www.kalva.demon.co.uk/soviet/ssoln/ssol616.html 

Related pages in this website

Summary of geometrical theorems

Ptolemy's Theorem -- Proof of the "equality" case

Puzzles

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