證明：?$(1)$?因?yàn)?$AD$?平分?$∠BAC，$??$BE$?平分?$∠ABC，$?
?$ $?所以?$∠BAE = ∠CAD，$??$∠ABE = ∠CBE。$?
?$ $?因?yàn)?$\overset {\frown }{CD}=\overset {\frown }{CD}，$?
根據(jù)同弧所對(duì)的圓周角相等，
所以?$∠DBC = ∠CAD，$?
?$ $?所以?$∠DBC = ∠BAE。$?
?$ $?因?yàn)?$∠DBE=∠CBE + ∠DBC，$??$∠DEB=∠ABE + ∠BAE，$?
?$ $?所以?$∠DBE=∠DEB，$?
根據(jù)等角對(duì)等邊，
所以?$DE = DB。$?
?$(2)$?連接?$CD。$?
?$ $?因?yàn)?$AD$?平分?$∠BAC，$?
所以?$\overset {\frown }{BD}=\overset {\frown }{CD}，$?
根據(jù)等弧對(duì)等弦，
所以?$CD = BD = 4。$?
?$ $?因?yàn)?$∠BAC = 90°，$?
所以?$BC$?是?$\triangle ABC$?外接圓的直徑?$(90°$?的圓周角所對(duì)的弦是直徑?$)，$?
?$ $?所以?$∠BDC = 90°。$?
?$ $?在?$Rt\triangle BDC$?中，
根據(jù)勾股定理?$BC=\sqrt {BD^2+CD^2}=\sqrt {4^2 + 4^2}=\sqrt {16 + 16}=\sqrt {32}=4\sqrt {2}。$?
?$ $?所以?$\triangle ABC$?外接圓的半徑?$=\frac {1}{2}BC=\frac {1}{2}×4\sqrt {2}=2\sqrt {2}。$?
