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Given that P is a point on the circum-circle (on arc AC) of an equilateral triangle ABC other than its vertices such that $$2PA = PC$$, then $$PA \colon PB$$ is
ABCP is a cyclic quadrilateral, so by Ptolemy's theorem $$PB \cdot AC = PA \cdot BC + PC \cdot AB$$. Since all sides of the equilateral triangle are equal, $$PB = PA + PC = PA + 2PA = 3PA$$. Therefore $$PA \colon PB = 1 \colon 3$$.
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