$解：(1)∵E是AB的中點$
$∴AE=\frac 12AB=1$
$在Rt△ADE中，由勾股定理，得$
$ED=\sqrt {{AE}^2+{AD}^2}=\sqrt {5}$
$∴AF=AH=EH-AE=ED-AE=\sqrt {5}-1，$
$DF=2-AF=3-\sqrt {5}$
$(2)是，\frac {AF}{AD}=\frac {\sqrt {5}-1}2，\frac {DF}{AF}=\frac {3-\sqrt {5}}{\sqrt {5}-1}=\frac {\sqrt {5}-1}2$
$∴\frac {AF}{AD}=\frac {DF}{AF}$
$∴F是AD的黃金分割點$
