(1)在Rt△ACD中,根据勾股定理有:AC 2 =AD 2 +DC 2 =a 2 +1 ∵∠AME=∠D=90°,∠EAM=∠CAD ∴△AME ∽ △ADC, ∴ AE AC = AM AD , ∴AE= AM•AC AD , ∵AM= 1 3 AC, ∴AE= a 2 +1 3a ;
(2)∵AE ∥ BC, ∴△AEM ∽ △CHM, ∴ AE CH = AM MC , ∵ AM AC = 1 3 , ∴ AE CH = 1 2 ,即CH=2AE= 2 a 2 +2 3a , ∴BH=a-CH= a 2 -2 3a , ∴ AE+BH a-AE+a-BH = 2 5 , ∴a 2 = 7 2 ,即a=
14 2 ;
(3)设AE=x, ∵AE ∥ BC, ∴ AM MC = AE BC , ∵ AM AC = 1 4 ,即 AM MC = 1 3 , ∴ AE BC = 1 3 , 设AE=x,则BC=3x,AC= 1+9 x 2 , ∵△AME ∽ △ADC, ∴ AE AC = AM AD , 由于AM= 1 4 AC,AD=BC, ∴x•3x= 1 4 (1+9x 2 ), ∴x=
3 3 , ∴AD=BC=3x= 3 ;
(4)由题意可知: AC= 1+ x 2 , AM= 1 4 1+ x 2 , ∵△AEM ∽ △ACD ∴ AE AC = AM AD ,∴AE= x 2 +1 4x , 同理可得出 AF AD = AE DC , ∴AF= x 2 +1 4 , 则S △AEF = 1 2 AE•AF= ( x 2 +1) 2 32x (