1.如图,已知矩形ABCD的对角线相交于点O,∠AOD=∠120°,过O作EF⊥BD,交AD于E,交BC于F,求证（1）EF=BF
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AO=DO=BO,
∠OAD=∠ODA=(180°-∠AOD)/2=(180°-120°)/2=30°,
AD//BC,∠ODA=∠CBD=30°,[内错角]
2OE=ED,
ED²=OE²+DO²
(2OE)²=OE²+DO²
OE²=DO²/3,
∠BOF=∠EOD=90°,∠CBD=30°
2OF=BF
OF²=BO²/3=DO²/3=EO²
OF=EO,
2OF=BF
OF+EO=BF
EF=BF.
[也可以用RT△EOD≌RT△FOB,(ASA)得EO=FO,FO=BF/2,FO=BF,EO+FO=BF,EF=BF]