解：（1）∵在梯形ABCD中,AD∥BC,AB=DC=AD=3,角ABC=60°∠ABC=60°∴∠EBC=∠AEB ∠A=∠D∴∠AEB+∠ABE=60°又∵∠BEF=120°∴∠AEB+∠DEF=60°∴∠ABE=∠DEF∴△ABE∽△DEF∴AB：DE=AE：DF即3：（3-x）=x：y∴y=x-x^2/3 0
如图,在梯形ABCD中,AD平行BC,AB=DC=AD=3,角ABC=60°
Simone
发布于 2024-01-16 12:54:55
564 阅读
