Question

E是正方形一边AD上一点,过上BE一点O作垂线,交AB于点G,交CD于点H,求证BE=GH

Answer 1

过点H做HM⊥AB,∠GHM+∠HGM=90,∠ABE+∠HGM=90,∠ABE=∠AHM

HM=AB,△GHM≌△ABE,GH=BE