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.

Answer 1

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.