ABC is an equilateral tiangle.P is any point on the arc BC.Prove AP=BP+PC

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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.**

**Sources:**

the link is this
http://www.flickr.com/photos/78780315@N06/7963176188/in/
photostream

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