矩形ABCD中,E为BC上一点,DF⊥AE于点F.(1)求证:△ABE∽△DFA;(2)若AB=6,AD=12,AE=10,求DF的长

2025-06-22 09:22:59
推荐回答(1个)
回答1:

解答:(1)证明:∵四边形ABCD为矩形,
∴AD∥BC,
∴∠AEB=∠DAF,
∵DF⊥AE,
∴∠B=∠AFD=90°,
∴△ABE∽△DFA;
(2)解:由(1)可知△ABE∽△DFA,

AB
DF
=
AE
AD

∵AB=6,AD=12,AE=10,
6
DF
=
10
12

解得DF=7.2.