摘要: (1) ∵ BD∥AC.点B.A.E在同一条直线上. ∴ ÐDBA = ÐCAE, 又∵ , ∴ △ABD∽△CAE. --- 4分 (2) ∵AB = 3AC = 3BD.AD =2BD . ∴ AD2 + BD2 = 8BD2 + BD2 = 9BD2 =AB2. ∴ÐD =90°, 由(1)得 ÐE =ÐD = 90°, ∵ AE=BD , EC =AD = BD , AB = 3BD . ∴在Rt△BCE中.BC2 = (AB + AE )2 + EC2 = (3BD +BD )2 + (BD)2 = BD2 = 12a2 , ∴ BC =a . --- 6分

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