# Ptolemy on Hinges

Ptolemy's theorem tells us that in a convex cyclic quadrilateral ABPC the sum of the products of the two pairs of opposite sides equals the product of its two diagonals. In other words,

AC×BP + AB×CP = AP×BC

The applet below illustrates a proof of Ptolemy's theorem that grew out of its application in a proof of a theorem by van Schooten. The proof is available online and is due to a former correspondent.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Rotate triangle ACP around A through the angle CAB so that C is mapped onto a point, say C', on AB. P is mapped to P'.

In case ΔABC is equilateral, one may perceive the semblance with the configuration from a proof of van Schooten's theorem. When ΔABC is equilateral the rotation places P' on the extension of BP. The same is true when ΔABC is isosceles. In the general case, we may only claim that C'P'||BP. (The reason is the same in all three cases. Since the quadrilateral ABPC is cyclic ∠ACP + ∠ABP = 180°. After the rotation, ∠AC'P' = ∠ABP'', where P'' is the intersection of AP' and BP.)

Triangles ABP'' and AC'P' are similar, implying

AC' / C'P' = AB / BP'',

so that

 BP'' = AB×C'P' / AC' = AB×CP / AC.

Triangles ABC and AP''P are also similar because, in addition to ∠PAP'' = ∠CAB, ∠'AP''P = ∠AP'C' = ∠APC = ∠ABC (the latter as the two inscribed angles subtended by the same arc.) This gives

AC / AP = BC / PP''

so that

PP'' = BC×AP / AC

But PP'' = BP + BP'' = BP + AB×CP / AC. We see that

BP + AB×CP / AC = BC×AP / AC.

Multiplying by AC gives Ptolemy's identity.

### References

1. A. Percy and D. G. Rogers, Alternative route: from van Schooten to Ptolemy, Normat 57:3, 116-128 (2009) ### Ptolemy's Theorem 