正五边形ABCDE中,AC与BE相交于点F.求证：四边形CDEF是菱形.
问题描述：
正五边形ABCDE中,AC与BE相交于点F.求证：四边形CDEF是菱形.
答
∠D=∠AED=∠BAE=∠CBA=∠BCd=108°
AB=AE
∠AEB=36°
∠FED=72°
∠BAC=36°
∠EAF=∠EFC=72°
AE=EF=DE
∠EFC=108°∠FCD=72°
∠FED=∠FCD=72°
∠EFC=∠D=108°
四边形CDEF是平行四边形
EF =DE
是菱形