Answer:
\[\overset{\to
}{\mathop{AB}}\,+\overset{\to }{\mathop{AC}}\,+\overset{\to
}{\mathop{AD}}\,+\overset{\to }{\mathop{AE}}\,+\overset{_{\to }}{\mathop{AF}}\,\]
\[=\overset{\to
}{\mathop{AB}}\,+(\overset{\to }{\mathop{AC}}\,+\overset{\to
}{\mathop{DC}}\,)+\overset{\to }{\mathop{AD}}\,+(\overset{\to
}{\mathop{AD}}\,+\overset{\to }{\mathop{DE}}\,)+\overset{_{\to
}}{\mathop{AF}}\,\]
\[=\overset{\to
}{\mathop{3AD}}\,+(\overset{\to }{\mathop{AB}}\,+\overset{\to
}{\mathop{DE}}\,)+(\overset{\to }{\mathop{DC}}\,+\overset{\to
}{\mathop{AF}}\,)=3\overset{_{\to }}{\mathop{AD}}\,\]
\[=3\times
(2\overset{\to }{\mathop{AO}}\,)=6\overset{\to }{\mathop{AO}}\,\]
[\[\because
\,\,\,\,\overset{\to }{\mathop{AB}}\,=-\overset{\to }{\mathop{DE}}\,\] and \[\overset{\to
}{\mathop{DS}}\,=-\overset{\to }{\mathop{AF}}\,\]]
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