过点B作BD⊥AC,交CA延长线于点D.则BD就是等腰△ABC的腰AC上的高；已知,AB = AC ,可得：∠C = ∠B = 15° ,∠BAD = ∠B+∠C = 30° ；因为,Rt△ABD中,∠ADB = 90° ,∠BAD = 30° ,所以,BD = AB/2 = a ；即有：等腰△...