證明：?$(1)$?因為?$D$?是?$\overset {\frown }{AC}$?的中點，
所以?$\overset {\frown }{AD}=\overset {\frown }{CD}。$?
?$ $?因為?$AB\perp DH，$?且?$AB$?是?$\odot O$?的直徑，
所以?$\overset {\frown }{AD}=\overset {\frown }{AH}，$?
則?$\overset {\frown }{CD}=\overset {\frown }{AH}。$?
根據(jù)同弧所對的圓周角相等，可得?$∠ADH = ∠CAD，$?
所以?$AF = DF。$?
?$ (2) $?設(shè)?$AE = \sqrt {5}x，$?因為?$\frac {AE}{AD}=\frac {\sqrt {5}}{5}，$?
所以?$AD = 5x。$?
?$ $?因為?$DE\perp AB，$?所以?$∠AED = 90°。$?
?$ $?在?$Rt\triangle AED$?中，根據(jù)勾股定理?$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。$?
?$ $?因為?$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。$?
展開式子：?$5x^2+20x^2-10\sqrt {5}x+\frac {25}{4}=\frac {25}{4}。$?
移項合并同類項：?$25x^2-10\sqrt {5}x = 0，$?
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。$?
