如图,四边形EFGH是矩形ABCD的内接矩形,且EF:FG=3:1,AB:BC=2:1,则tan∠AHE的值为(  ) A.15 B.310 C.16 D.27

问题描述:

如图,四边形EFGH是矩形ABCD的内接矩形,且EF:FG=3:1,AB:BC=2:1,则tan∠AHE的值为(  )
A.

1
5

B.
3
10

C.
1
6

D.
2
7

∵四边形EFGH是矩形ABCD的内接矩形,EF:FG=3:1,AB:BC=2:1,
∴∠HEA+∠FEB=90°,
∵∠FEB+∠EFB=90°,
∴∠HEA=∠EFB,
∵∠HAE=∠B,
∴Rt△HAE∽△EBF,

HA
EB
=
AE
FB
=
HE
EF
=
1
3

同理可得,∠GHD=∠EFB,HG=EF,
∴△GDH≌△EBF,DH=BF,DG=EB,
设AB=2x,BC=x,AE=a,BF=3a,
则AH=x-3a,AE=a,
∴tan∠AHE=tan∠BEF,
a
x−3a
=
3a
2x−a
,解得:x=8a,
∴tan∠AHE=
a
x−3a
=
a
8a−3a
=
1
5

故选A