延长AD,BC相交于E.
∵CD平行AB,∴∠ABC=∠DCE.
∵∠ADC=∠AEB+∠ECD=2∠ABC=2∠ECD,
∴∠ECD=∠AEB=∠ABC.
∴CD=DE=AE-AD=AB-AD=b-a.
故答案是:b-a.
如图,四边形ABCD中,AB∥CD,∠D=2∠B,若AD=a,AB=b,则CD的长是_.
如图,四边形ABCD中,AB∥CD,∠D=2∠B,若AD=a,AB=b,则CD的长是______.
数学人气:267 ℃时间:2020-01-27 16:23:26
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