Question

已知：角A+角C=角AEC,求证：AB平行CD

Answer 1

证明：过点E作FE‖AB,

∴∠AEF=∠A,

∵∠AEC=∠A+∠C,

即 ∠AEF+∠CEF=∠A+∠C.

∴∠CEF=∠C.

∴EF‖CD,

∴AB‖EF,CD‖EF.

∴AB‖CD.