题目内容

如图,AB = 3ACBD = 3AE,又BDAC,点BAE在同一条直线上.

(1) 求证:△ABD∽△CAE

(2) 如果AC =BDAD =BD,设BD = a,求BC的长.

 (1) ∵ BDAC,点BAE在同一条直线上,   ∴ ÐDBA = ÐCAE,

又∵ , ∴  △ABD∽△CAE.                              

(2)  ∵AB = 3AC = 3BDAD =2BD

(第22题)

 ∴ AD2 + BD2 = 8BD2 + BD2 = 9BD2 =AB2,               

∴ÐD =90°,             

由(1)得 ÐED = 90°, 

AE=BD , EC =AD = BD , AB = 3BD

∴在Rt△BCE中,BC2 = (AB + AE )2 + EC2

= (3BD +BD )2 + (BD)2 = BD2 = 12a2 ,

(第23题)

  ∴ BC =a .                                    

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