如图,已知梯形ABCD中,AD∥CB,E,F分别是BD,AC的中点,BD平分∠ABC.
已知:如图,在梯形ABCD中,AD∥BC,AB=AD,∠BAD的平分线AE交BC于点E,连接DE.求证:四边形ABED是菱形.
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如图,梯形ABCD中,AB∥CD,∠ABC=90°,AD=DC+AB,DE=DC,F为BC中点.| 1 |
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线段BE上取一点G,使得BF=BG,连接CG.| 2 |
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如图,梯形ABCD中,AD∥BC,点E在BC上,AE=BE,且AF⊥AB,连接EF.
如图,已知四边形ABCD是梯形,AD∥BC,∠A=90°,BC=BD,CE⊥BD,垂足为E.