Question

一个四边形ABCD,E,F分别在BC和CD上,AB=AE=AF=AD=BC=CD=EF,求角C的度数.

Answer 1

连接A、C,易证AC⊥EF

所以∠ACF+∠EFC=90°

而∠EFC+∠AFD=180°-60°=120°

所以∠D=∠AFD=∠ACF+30°

∠DAC+∠ACF+∠D=∠ACF+∠ACF+∠ACF+30°=180°

∠ACF=50°

∠C=2∠ACF=100°