Question

已知梯形ABCD,AD//BC,对角线AC,BD相交于点E,AB=AC,并且AB垂直于AC,BD=BC,证明CD=CE

Answer 1

过B作BE⊥DA交DA的延长线于E.
∵AB⊥AC,AB＝AC,∠ABC＝∠ACB＝45°,且AB＝BC/√2.
∵AD∥BC,BE⊥AD,∴BE⊥BC,∴∠ABE＝90°－∠ABC＝90°－45°＝45°.
∴BE＝AB/√2＝（BC/√2）/√2＝BC/2,∴∠ADB＝30°.
∵AD∥BC,∴∠CBD＝∠ADB＝30°.
∴∠BCD＋∠BDC＝180°－∠CBD＝180°－30°＝150,∴∠CDE＝75°.
由三角形外角定理,有：∠CED＝∠CBD＋∠ACB＝30°＋45°＝75°.
由∠CDE＝75°、∠CED＝75°,得：∠CDE＝∠CED,∴CD＝CE.