△ABC外切⊙O于D,E,F三点,内切园半径为1,∠C=60°,AB=5,求△ABC的周长和面积

Question

△ABC外切⊙O于D,E,F三点,内切园半径为1,∠C=60°,AB=5,求△ABC的周长和面积

Answer 1

连接OF,OE,OD∵△ABC外切⊙O于D,E,F三点∴OE=OF=ODCE=CF,AD=AF,BD=BE∴△OFC全等于△OEC△ADO全等于△AFO△BDO全等于△BEO∴∠OCE=∠OCF=30°∴CE=CF=根号3∴△ABC的周长是AB+BC+AC=5+5+2根号3=10+2√3S△ABC=2S△ADO+2S△BDO+2S△CEO=2S△ABO+2S△CEO=5/2+√3/2=(5+√3)/2