B
四边形CDC′E是菱形.
理由:根据折叠的性质,可得:CD=C′D,∠C′DE=∠CDE,CE=C′E,
∵AD∥BC,
∴∠C′DE=∠CED,
∴∠CDE=∠CED,
∴CD=CE,
∴CD=C′D=C′E=CE,
∴四边形CDC′E为菱形.
故选B.
B
四边形CDC′E是菱形.
理由:根据折叠的性质,可得:CD=C′D,∠C′DE=∠CDE,CE=C′E,
∵AD∥BC,
∴∠C′DE=∠CED,
∴∠CDE=∠CED,
∴CD=CE,
∴CD=C′D=C′E=CE,
∴四边形CDC′E为菱形.
故选B.