$解：因为C为AB的黄金分割点，AC > BC$
$所以AC²= AB×BC= AB×(AB-AC)$
$所以\frac {AC}{AB}=\frac {AB-AC}{AC}=\frac {AB}{AC}-1$
$设\frac {AC}{AB}=x$
$则x=\frac {1}{x}-1$
$所以解得x=\frac {\sqrt{5}-1}{2}或x=\frac {-\sqrt{5}-1}{2}(舍去)$
$所以\frac {AC}{AB}=\frac {\sqrt{5}-1}{2}$
