C
分析:根據切線長定理可以證得:BF+CH=BG+CG=BC,DE=DR+ER=DF+EH,根據△ADE的周長=AD+AE+DE=AD+AE+DF+EH=AF+AH=△ABC的周長-(BF+CH)=△ABC的周長-BC即可求解.
解答:
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解:∵⊙I是△ABC的內切圓,
∴BF=BG,CG=CH,DR=DF,ER=EH
∴BF+CH=BG+CG=BC=5,
DE=DR+ER=DF+EH,
∴△ADE的周長=AD+AE+DE=AD+AE+DF+EH=AF+AH=△ABC的周長-BC-(BF+CH)=△ABC的周長-2BC=20-2×5=10.
故選C.
點評:本題考查了切線長定理,正確理解∴△ADE的周長=AD+AE+DE=AD+AE+DF+EH=AF+AH=△ABC的周長-(BF+CH)=△ABC的周長-BC是關鍵.