如图,在平行四边形ABCD中,∠C=60°,M、N分别是AD、BC的中点,BC=2CD.

如图,在平行四边形ABCD中,∠C=60°,MN分别是ADBC的中点,BC=2CD.

  (1)求证:四边形MNCD是平行四边形;

  (2)求证:BD=3MN.

答案

证明:(1)ABCD是平行四边形,

AD=BCADBC.

MN分别是ADBC的中点,

MD=NCMDNC.

MNCD是平行四边形;

  (2)连接ND

MNCD是平行四边形,

MN=DC.

NBC的中点,

BN=CN.

BC=2CD,∠C=60°,

∴△NCD是等边三角形.

ND=NC,∠DNC=60°.

∵∠DNC是△BND的外角,

∴∠NBD+NDB=DNC.

DN=NC=NB

∴∠DBN=BDN=DNC=30°.

∴∠BDC=90°.

BC=2DCBD===DC.

DC=MN,∴BD=MN.

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