Given a equilateral triangle ABC and a point P on the short chord {BC} in the circumcircle of the triangle. Prove that: AP = BP + CP
From Ptolemy’s Theorem, we will get:
AP*BC=BP*AC+CP*AB
Since AB=BC=AC
Then AP=BP+PC
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Take the point X on AP such that PX = PC.
AP = BP + CP \iff AX = BP.
We observe that
\angle XPC = \angle APC = \angle ABC = 60^\circ.
Since PC = PX, it follows that
PC = CX = XP.
Moreover,
\angle PCX = 60^\circ = \angle ACB \ \Rightarrow \ \angle BCP = \angle ACX.
Now we have
AC = BC, \ \angle BCP = \angle ACX, \ PC = XC,
Therefore,
\triangle ACX \cong \triangle BCP. (SAS)
Hence,
AX = BP.
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