Question

如图,梯形ABCD中,AD平行BC,AD大于BC,角B+角C=90,E为AD中点,F为BC中点,求证,

求证EF=1/2(BC-AD)

Answer 1

过F做FG//AB,FH//CD,分别交AD于G，H

∵FG//AB
∴∠B+∠BFG = 180°
∵FH//CD
∴ ∠C+∠CFD = 180°
∵∠B+∠C = 90°
∴∠BFG+∠CFD = 90°
∵AD//BC
∴∠BFG = ∠DGF
∠CFD = ∠AHF
∴∠HGF+∠GHF = 90°
∵F是BC的中点
∴BF = CF
∵AD//BC，FG//AB,FH//CD
∴AG = BF = CF = HD
∵E是AD的中点
∴AE = ED
∴GE = EH
∴E是GH的中点
∴EF是Rt△GFH的斜边GH的中线
∴2EF = GH
∵GH = AD - BC
∴EF=(BC-AD)/2