Question

如图,在梯形ABCD中,AD‖BC,∠B+∠C=90°,点E、F分别是AD,BC中点,AD=4,BC=10,求EF的长

Answer 1

过E做EG//AB,EH//DC∵∠B+∠C=90°∠B = ∠EGH,∠C =∠EHG∴∠EGH+∠EHG = 90°∴∠GEH=90°∵F是BC的中点∴BG+GF = FH+CH∵E是AD的中点∴AE =ED∵AE = BG,ED = CH∴GF = HF∴F是GH的中点∴EF是Rt△EGH的中线∴EF = GM/2 = (BC-AD)/2 = 3