Question

如图,AD‖BC,AD=BC,AB=2BC,E为DC的中点,AE与BC延长相交于点F。求证:∠F=

∠FAB

Answer 1

已知AD//BC，AD=BC
则四边形ABCD为平行四边形
AB∥CD，CD=AB
AB=2BC，则DC=2AD
E为CD中点，ED=AD
∠EAD=∠AED

AD//BC，则∠F=∠EAD=∠AED
∠AED与∠CEF为对顶角
∠AED=∠CEF
∠F=∠CEF

AB∥CD，则∠FAB=∠CEF

得∠F=∠FAB