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Question
ABCDE is a pentagon, prove that
\[\overrightarrow{AB} + \overrightarrow{AE} + \overrightarrow{BC} + \overrightarrow{DC} + \overrightarrow{ED} + \overrightarrow{AC} = 3\overrightarrow{AC}\]
Solution
Given: ABCDE is a pentagon.
To Prove: \[\overrightarrow{AB} + \overrightarrow{AE} + \overrightarrow{BC} + \overrightarrow{DC} + \overrightarrow{ED} + \overrightarrow{AC} = 3 \overrightarrow{AC} .\]
Proof: We have,
\[LHS = \overrightarrow{AB} + \overrightarrow{AE} + \overrightarrow{BC} + \overrightarrow{DC} + \overrightarrow{ED} + \overrightarrow{AC} \]
\[= \left( \overrightarrow{AB} + \overrightarrow{BC} \right) + \left( \overrightarrow{AE} + \overrightarrow{ED} \right) + \overrightarrow{DC} + \overrightarrow{AC}\]
\[= \overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{DC} + \overrightarrow{AC}\] [∵ \[\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}\] and \[\overrightarrow{AE} + \overrightarrow{ED} = \overrightarrow{AD}\]]
\[\overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AC} - \overrightarrow{AD} + \overrightarrow{AC}\] [∵ \[\overrightarrow{AD} + \overrightarrow{DC} = \overrightarrow{AC} \Rightarrow \overrightarrow{DC} = \overrightarrow{AC} - \overrightarrow{AD} \]]
\[= 3 \overrightarrow{AC} = RHS\]
Hence proved.
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