ABC is an equilateral tiangle.P is any point on the arc BC.Prove AP=BP+PC http://www.flickr.com/photos/78780315@N06/7963176188/in/photostream
Since the points ABPC lie on a common circle, then the quadrilateral is cyclic (see link below). Cyclic quadrilaterals obey Ptolemy's theorem, so we have :
`AP cdot BC= AC cdot BP+PC cdot AB`
but since triangle ABC is equilateral, each side is the same. Let
then Ptolemy's Theorem becomes
`AP cdot x=x cdot BP+PC cdot x` divide both sides by x to get
The theorem has been proven.