Question

如图,在菱形ABCD中,AE⊥BC,垂足为E,AF⊥CD,垂足为F,且BE=EC,CF=DF

Answer 1

连接AC

∵四边形ABCD是菱形

∴BC=DC=AB=AD,∠D=∠B,∠BAD=∠BCD

又∵BE=CE,CF=DF

∴E是BC的中点,F是DC的中点

∴DF=CF=CE=EB

又∵AE⊥BC,AF⊥DC

∴∠AFD=∠AFC=∠AEB=∠AEC=90°

∴△AFD≌△AFC≌△AEB≌△AEC

∴∠ADF=∠ACF=∠ABE=∠ACE

又∵∠B+∠DCB=180°

∴3∠B=180°

∠B=60°

∴∠DCB=120°

∴∠EAF=60°