text: # 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
  
  lfryerda | Certified Educator
  
  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
  
  `x=AB=AC=BC`
  
  then Ptolemy's Theorem becomes
  
  `AP cdot x=x cdot BP+PC cdot x`   divide both sides by x to get
  
  `AP=BP+PC`
  
  The theorem has been proven.
  
  topperoo | Student
  the link is this http://www.flickr.com/photos/78780315@N06/7963176188/in/ photostream