Question

梯形ABCD中,AD‖BC,AD⊥DC,E是CD的中点,AE比BE=DE比BC,求证:AE平分∠BAD

Answer 1

△ADE∽△BCE,∴∠DAE=∠BEC∵∠DAE+∠AED=90°,∴∠BEC+∠AED=90°,∴∠AEB=90°.

过点E作EF∥BC交AB于F,则EF为梯形中位线.∴F为AB中点.在直角△AEB中,EF=1/2AB=FA,

∴∠AEF=∠FAE=∠EAD∴AE平分∠BAD