Question

如图,△ABC是等腰三角形,AB=AC,角A＝60°,D为AC边的中点,延长BC到点E,使CE=CD,连接BD、DE.若

Answer 1

∵△ABC是等腰△,∠A = 60°∴等腰△ABC为等边三角形∴∠ACB = 60°,AB = BC∵D是AC的中点∴BD是∠ABC 的角平分线(等腰三角形三线合一)∴∠DBC = 30°∵CD = CE∴2∠E = 2∠CDE = ∠ACB∴∠E = 30°∴BD = DE∵△ABC...