(2014?新乡一模)如图,已知?ABCD.(1)尺规作图:连接AC,作∠ABC的平分线BF分别与AC,AD交于点E,F;

2025-05-07 17:08:26
推荐回答(1个)
回答1:

(1)解:如图所示:

(2)证明:∵?ABCD中AD∥BC,
∴∠AFB=∠FBC,
又∵∠ABF=∠FBC,
∴∠ABF=∠AFB,
∴AB=AF;

(3)解:∵AD∥BC,
∴△AEF∽△CEB,
当AB=3,BC=5时,
∴AF=3,

AE
EC
=
AF
BC
=
3
5

AE
AC
的值为:
3
8