\(\overrightarrow{AM}\cdot\overrightarrow{BC}=\overrightarrow{BC}\left(\overrightarrow{BM}-\overrightarrow{BA}\right)=\overrightarrow{BM}\cdot\overrightarrow{BC}-\overrightarrow{BC}\cdot\overrightarrow{BA}\)
\(=BM\cdot BC\cdot cos0^0=\dfrac{1}{2}\cdot a^2\cdot1=\dfrac{1}{2}a^2\)
\(\left|\overrightarrow{AM}+\overrightarrow{BC}\right|=\sqrt{AM^2+BC^2+2\cdot\dfrac{1}{2}a^2}\)
\(=\sqrt{\dfrac{1}{4}a^2+a^2+a^2+a^2}=\dfrac{\sqrt{13}}{2}\cdot a\)
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