Question

直角梯形ABCD中,AD‖BC,角ADC=90度AB=5,CD=3,AD=8,线段AD上移动点E,过E做EF垂直AB

Answer 1

1) 作BB'垂直AD于B'

BB'=CD=3 , AB=5 => AB'=4

△AB'B中, BB':AB':AB = 3:4:5

EF垂直AB, BB'垂直AB',∠A=∠A

=>△AFE ∽ △AB'B

=>EF:AE = BB':AB =3:5

=>Y/(8-X) = 3/5 => 5Y = 3(8-X) => Y=3/5 (8-X)

2) △AEF ∽ △CED

=> EF:AF = ED:CD = BB':AB' = 3:4

=> X:3 = 3:4

=> X = 9/4