# A nice, unsolved geometry problem

• Feb 16th 2010, 12:33 PM
oswaldo
A nice, unsolved geometry problem
• Feb 18th 2010, 08:11 AM
BobP
There is a theorem, (easily proved using similar triangles), involving the intersection of two chords in a circle.
If as a result of the intersection, one chord splits into lengths a and b, while the other splits into lengths p and q, then ab = pq.
Apply this at each of the four corners of the rhombus, add the four equations, cancel this down and you get the required equation.
• Feb 22nd 2010, 11:55 AM
oswaldo
I think the theorem which states: "ab = pq" must be Ptolemy. But I am not sure about its name.
• Feb 23rd 2010, 01:20 AM
BobP
No, this is not Ptolemy's theorem.
For this one, put two intersecting chords in a circle. Call them say AB and PQ and call the point of intersection C.
By the angles on the same arc being equal theorem you have a choice of two sets of similar triangles, ACQ and PCB, or ACP and BCQ.
Choose one of the two, equate corresponding ratios, cross multiply and you have your result.