∵∠BDC＝60°∴∠ADB＝120°∵∠A＝45°∴∠ABD＝15°∵BD评分∠ABC∴角ABD＝∠CBD＝15°∵DE平行于CB∴∠CBD＝∠BDE＝15°∴∠BED＝150°2.∵∠BAC为三角形AFG外角∴∠AFG加∠G＝∠BAc又∵AD为△AbC的平分线∴∠ABD等于∠CAD∴∠G＝∠CAD∴GE平行于AD
