证明：$(1)$因为$D$是$\overset {\frown }{AC}$的中点，
所以$\overset {\frown }{AD}=\overset {\frown }{CD}。$
$ $因为$AB\perp DH，$且$AB$是$\odot O$的直径，
所以$\overset {\frown }{AD}=\overset {\frown }{AH}，$
则$\overset {\frown }{CD}=\overset {\frown }{AH}。$
根据同弧所对的圆周角相等，可得$∠ADH = ∠CAD，$
所以$AF = DF。$
$ (2) $设$AE = \sqrt {5}x，$因为$\frac {AE}{AD}=\frac {\sqrt {5}}{5}，$
所以$AD = 5x。$
$ $因为$DE\perp AB，$所以$∠AED = 90°。$
$ $在$Rt\triangle AED$中，根据勾股定理$DE=\sqrt {AD^2-AE^2}=\sqrt {(5x)^2-(\sqrt {5}x)^2}$
$=\sqrt {25x^2 - 5x^2}=\sqrt {20x^2} = 2\sqrt {5}x。$
$ $因为$AF=\frac {5}{2}，$$AF = DF，$
所以$DF=\frac {5}{2}。$
$ $在$Rt\triangle AEF_{中}，$$EF=DE - DF=2\sqrt {5}x-\frac {5}{2}。$
$ $由$AE^2+EF^2=AF^2，$可得$(\sqrt {5}x)^2+(2\sqrt {5}x - \frac {5}{2})^2=(\frac {5}{2})^2。$
展开式子：$5x^2+20x^2-10\sqrt {5}x+\frac {25}{4}=\frac {25}{4}。$
移项合并同类项：$25x^2-10\sqrt {5}x = 0，$
提取公因式$5x$得$5x(5x - 2\sqrt {5}) = 0。$
$ $解得$x_{1}=0($舍去$)，$$x_{2}=\frac {2\sqrt {5}}{5}。$
$ $所以$AE=\sqrt {5}x=\sqrt {5}×\frac {2\sqrt {5}}{5}=2。$
