问题描述：
平行四边形ABCD中,AM⊥BC于M,AN⊥CD于N,已知AB=10,BM=6,MC=3.求线段MN的长
问题解答：
我来补答
∵ABCD是平行四边形
∴BC=AD=BM+CM=6+3=9
AB=CD=10
∠B=∠D
∵AM⊥BC,AN⊥CD
∴AM²=AB²-BM²=10²-6²=8²,AM=8
△ABM∽△ADN
AB/AD=AM/AN=BM/DM,AN=AD×AM/AB=9×8/10=36/5
DN=AD×BM/AB=9×6/10=27/5
∴CN=CD-DN=10-27/5=23/5
∵cosB=BM/AB=6/10=3/5
∴cosC=cos(180°-B)=-cosC=-3/5
∴MN²=CM²+CN²-2CM×CN×cosC
=3²+(23/5)²+2×3×23/5×3/5
=1168/25
MN=4√73/5
∴BC=AD=BM+CM=6+3=9
AB=CD=10
∠B=∠D
∵AM⊥BC,AN⊥CD
∴AM²=AB²-BM²=10²-6²=8²,AM=8
△ABM∽△ADN
AB/AD=AM/AN=BM/DM,AN=AD×AM/AB=9×8/10=36/5
DN=AD×BM/AB=9×6/10=27/5
∴CN=CD-DN=10-27/5=23/5
∵cosB=BM/AB=6/10=3/5
∴cosC=cos(180°-B)=-cosC=-3/5
∴MN²=CM²+CN²-2CM×CN×cosC
=3²+(23/5)²+2×3×23/5×3/5
=1168/25
MN=4√73/5
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