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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
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.
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