如图:在平行四边形ABCD的边AB,CD上截取AF,CE,使得AF=CE,连接EF,

如图:在平行四边形ABCD的边ABCD上截取AFCE,使得AF=CE,连接EF,点MN是线段EF上两点,且EM=FN,连接ANCM

1)求证:△AFN≌△CEM

2)若∠CMF=107°,∠CEM=72°,求∠NAF的度数.

答案

1)证明:∵四边形ABCD是平行四边形,

CDAB

∴∠AFN=CEM

FN=EMAF=CE

∴△AFN≌△CEMSAS).

2)解:∵△AFN≌△CEM

∴∠NAF=ECM

∵∠CMF=CEM+ECM

107°=72°+ECM

∴∠ECM=35°

∴∠NAF=35°

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