过点F、C、G分别橘孝型做AB的垂线,垂足为P、Q、R则CQ为梯形FPRG中位线设AD=2a,BD=2b易知AP=PD=aDR=RB=bAE=EB=a+b所以QE=BE-BD=a+b-2b=a-bQD=AD-AE=2a-(a+b)=a-b所以Q为ED中点,则CQ垂直平分ED,圆猜CD=CE望采纳!!!!!!慎此!!!!!!