Question

如图,在梯形ABCD中,AD//BC,AB=DC=AD=6,∠ABC=60°,点E在AD边上,∠BEF=120°,设AE

Answer 1

⑴∵AD∥BC,AB=CD,∠ABC=60°,∴∠A=∠D=120°,

∴∠ABE+∠AEB=60°,

∵∠BEF=120°,∴∠AEB+∠EDF=60°,

∴∠ABE=∠DEF,

∴ΔABE∽ΔDEF,∴AB/DE=AE/DF,

6/(6-X)=X/Y,

Y=-1/6X^2+X.

⑵由⑴相似得：EF/BE=DF/AE=Y/X,

∵ΔABE∽ΔEBF,

∴EF/BE=AE/AB=X/6,

∴Y/X=X/6,Y=1/6X^2,

∴-1/6X^2+X=1/6X^2,

X^2-3X=0

X=3或0(舍去),

∴E为AD的中点.