Question

AB=3AC，BD=3AE，又BD平行AC，点B，A，E在同一直线上

Answer 1

1、(2010年杭州市)如图,AB = 3AC,BD = 3AE,又BD‖AC,点B,A,E在同一条直线上.
(1) 求证:ABD∽CAE;
(2) 如果AC =BD,AD =BD,设BD = a,求BC的长.
答案:
(1) ∵ BD‖AC,点B,A,E在同一条直线上, ∴ (DBA = (CAE,
又∵ , ∴ ABD∽CAE.
(2) ∵AB = 3AC = 3BD,AD =2BD ,
∴ AD2 + BD2 = 8BD2 + BD2 = 9BD2 =AB2,
∴(D =90°
由(1)得(E =(D = 90°,
∵ AE=BD , EC =AD = BD , AB = 3BD ,
∴在RtBCE中,BC2 = (AB + AE )2 + EC2
= (3BD +BD )2 + (BD)2 = BD2 = 12a2 ,
∴ BC =a .

Answer 2

题目不全