如图,在直角梯形ABCD中,AB∥DC,∠ABC=90°,AB=2DC,对角线AC⊥BD,垂足为F,过点F作EF∥AB……急需!如图,在直角梯形ABCD中,AB∥DC,∠ABC=90°,AB=2DC,对角线AC⊥BD,垂足为F,过点F作EF∥AB,交AD于点E,CF=4cm.(1)求
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![如图,在直角梯形ABCD中,AB∥DC,∠ABC=90°,AB=2DC,对角线AC⊥BD,垂足为F,过点F作EF∥AB……急需!如图,在直角梯形ABCD中,AB∥DC,∠ABC=90°,AB=2DC,对角线AC⊥BD,垂足为F,过点F作EF∥AB,交AD于点E,CF=4cm.(1)求](/uploads/image/z/2768816-56-6.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E7%9B%B4%E8%A7%92%E6%A2%AF%E5%BD%A2ABCD%E4%B8%AD%2CAB%E2%88%A5DC%2C%E2%88%A0ABC%3D90%C2%B0%2CAB%3D2DC%2C%E5%AF%B9%E8%A7%92%E7%BA%BFAC%E2%8A%A5BD%2C%E5%9E%82%E8%B6%B3%E4%B8%BAF%2C%E8%BF%87%E7%82%B9F%E4%BD%9CEF%E2%88%A5AB%E2%80%A6%E2%80%A6%E6%80%A5%E9%9C%80%21%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E7%9B%B4%E8%A7%92%E6%A2%AF%E5%BD%A2ABCD%E4%B8%AD%2CAB%E2%88%A5DC%2C%E2%88%A0ABC%3D90%C2%B0%2CAB%3D2DC%2C%E5%AF%B9%E8%A7%92%E7%BA%BFAC%E2%8A%A5BD%2C%E5%9E%82%E8%B6%B3%E4%B8%BAF%2C%E8%BF%87%E7%82%B9F%E4%BD%9CEF%E2%88%A5AB%2C%E4%BA%A4AD%E4%BA%8E%E7%82%B9E%2CCF%3D4cm%EF%BC%8E%EF%BC%881%EF%BC%89%E6%B1%82)
如图,在直角梯形ABCD中,AB∥DC,∠ABC=90°,AB=2DC,对角线AC⊥BD,垂足为F,过点F作EF∥AB……急需!如图,在直角梯形ABCD中,AB∥DC,∠ABC=90°,AB=2DC,对角线AC⊥BD,垂足为F,过点F作EF∥AB,交AD于点E,CF=4cm.(1)求
如图,在直角梯形ABCD中,AB∥DC,∠ABC=90°,AB=2DC,对角线AC⊥BD,垂足为F,过点F作EF∥AB……急需!
如图,在直角梯形ABCD中,AB∥DC,∠ABC=90°,AB=2DC,对角线AC⊥BD,垂足为F,过点F作EF∥AB,交AD于点E,CF=4cm.
(1)求证:四边形ABFE是等腰梯形;
(2)求AB的长.
如图,在直角梯形ABCD中,AB∥DC,∠ABC=90°,AB=2DC,对角线AC⊥BD,垂足为F,过点F作EF∥AB……急需!如图,在直角梯形ABCD中,AB∥DC,∠ABC=90°,AB=2DC,对角线AC⊥BD,垂足为F,过点F作EF∥AB,交AD于点E,CF=4cm.(1)求
⑴过D点作AB的垂线,垂足为H点,则四边形DHBC是矩形,∴BH=DC,设DC=x,则HB=x,∴AH=x,∴H是AB中点,∴△DAB是等腰△,即DA=DB,又∵EF∥AB,则EA=FB,∴四边形ABFE是等腰梯形.⑵设DF=y,则由DC∥AB得:△DCF∽△BAF,∴BF=2y,CF=4,则AF=8,∴由勾股定理得:①CF²+BF²=CB²=DB²-HB²,即:4²+﹙2y﹚²=﹙3y﹚²-x²,整理得:5y²-x²=16,②DF²+CF²=DC²,即y²+4²=x²,联合①②解得:x=2√6,y=2√2,∴AB=2x=4√6