??$解：因為C為AB的黃金分割點，AC > BC$??
??$所以AC2= AB×BC= AB×(AB-AC)$??
??$所以\frac {AC}{AB}=\frac {AB-AC}{AC}=\frac {AB}{AC}-1$??
??$設(shè)\frac {AC}{AB}=x$??
??$則x=\frac {1}{x}-1$??
??$所以解得x=\frac {\sqrt{5}-1}{2}或x=\frac {-\sqrt{5}-1}{2}(舍去)$??
??$所以\frac {AC}{AB}=\frac {\sqrt{5}-1}{2}$?
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