1、(1) X;(2) X;(3) √;(4) X.
2、3,AD、CD、AE.
3、8.
4、∠AOB = 60°;过点O作AB的垂线,垂足
为C,在Rt△AOC中,AC = 1/2AB = 3,
OA = 6,
.
5、(1)△DOE为等边三角形;
(2)由题意,得∠B + ∠C = 120°.由OB =
OD,OC = OE,得∠ODB = ∠B,
∠OEC = ∠C,所以∠ODB + ∠OEC = 120°,∠BOD + ∠COE = 120°,∠DOE = 60°.
又OD = OE,因此△DOE为等边三角形.