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

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lfryerda's profile pic

Posted on

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.

topperoo's profile pic

Posted on

the link is this photostream

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