Question

如图,在梯形ABCD中,AD∥BC,AB=DC,过点D作DE⊥BC,垂足为E,并延长DC至点F,使EF=DE,连接BF,

Answer 1

（1）证明：∵DF⊥BC,DE = DF

∴BC垂直平分DF

∴DC = CF = AB∠DCB =∠FCB =∠ABC

/ a>∴AB平行CF

AB平行CF AB = CF.

（2）连接到BD.

DE：EC = 3:2 = BE：DE

另一个∵∠DEC =∠DEB = 90°

∴△DEC△BED类似

>

∴∠EDC =∠DBC

∴∠BDC =∠EDC +∠BDE =∠DBC +∠BDE = 90°

∴∠BAC =∠ BDC = 90°

∵的四边形ABFC平行四边形

∴四边形ABFC矩形.