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Question
If P is a point and ABCD is a quadrilateral and \[\overrightarrow{AP} + \overrightarrow{PB} + \overrightarrow{PD} = \overrightarrow{PC}\], show that ABCD is a parallelogram.
Solution
Given: ABCD is a quadrilateral such that \[\overrightarrow{AP} + \overrightarrow{PB} + \overrightarrow{PD} = \overrightarrow{PC} .\]
To show: ABCD is a parallelogram.
Proof: Consider,
\[\overrightarrow{AP} + \overrightarrow{PB} + \overrightarrow{PD} = \overrightarrow{PC} \]
\[ \Rightarrow \overrightarrow{AP} + \overrightarrow{PB} = \overrightarrow{PC} - \overrightarrow{PD}\]
\[\Rightarrow \overrightarrow{AB} = \overrightarrow{DC}\] [ ∵ \[\overrightarrow{AP} + \overrightarrow{PB} = \overrightarrow{AB}\] and \[\overrightarrow{PD} + \overrightarrow{DC} = \overrightarrow{PC}\]]
Again,
\[\overrightarrow{AP} + \overrightarrow{PB} + \overrightarrow{PD} = \overrightarrow{PC} \]
\[ \Rightarrow \overrightarrow{AP} + \overrightarrow{PD} = \overrightarrow{PC} - \overrightarrow{PB}\]
\[\Rightarrow \overrightarrow{AD} = \overrightarrow{BC}\] [ ∵ \[\overrightarrow{AP} + \overrightarrow{PD} = \overrightarrow{AD}\] and \[\overrightarrow{PB} + \overrightarrow{BC} = \overrightarrow{PC}\]}
Since, opposite sides of the quadrilateral are equal and parallel.
Hence,
ABCD is a parallelogram.
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