在平行四邊形ABCD中,AB=2BC,E為BA中點(diǎn),DF垂直BC,垂足為F,證明∠AED=∠EFB
在平行四邊形ABCD中,AB=2BC,E為BA中點(diǎn),DF垂直BC,垂足為F,證明∠AED=∠EFB
數(shù)學(xué)人氣：136 ℃時(shí)間：2020-01-31 22:38:53
優(yōu)質(zhì)解答
證明:過E點(diǎn)作BC的平行線交DF于M,即EM‖BC∵E是AB的中點(diǎn)∴EM是梯形BFDA的中位線∴M是DF的中點(diǎn)∵DF⊥BC∴EM⊥DF∴三角形DEF為等腰三角形∴∠EDF=∠EFD又因?yàn)锳D‖BC且DF⊥BC∴∠ADF=∠BFD=90°∵∠ADE=∠ADF-∠EDF,∠EF...
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