Question

ABCD are four points in a plane and Q is the point of intersection of the lines joining the mid-points of AB and CD; BC and AD. Show that\[\vec{PA} + \vec{PB} + \vec{PC} + \vec{PD} = 4 \vec{PQ}\], where P is any point.

Answer 1

Let E, F, G and H are the midpoints of the sides AB, BC, CD and DA respectively of say quadrilateral ABCD.  By geometry of the figure formed by joining the midpoints E, F, Gand H will be a parallelogram. Hence its diagonals will bisect each other, say at Q.
Now, F is the midpoint of BC.
\[\frac{\overrightarrow{PB}+ \overrightarrow{PC}}{2} = \overrightarrow {PF} \]
\[ \therefore\overrightarrow {PB} +\overrightarrow {PC} = 2 \overrightarrow {PF} . . . . . \left( 1 \right)\]
And, H is the midpoint of AD.
\[\frac{\overrightarrow {PB} + \overrightarrow {PC}}{2} =\overrightarrow {PF} \]
\[ \therefore\overrightarrow {PB} + \overrightarrow{PC} = 2 \overrightarrow {PF} . . . . . \left( 1 \right)\]
Adding (1) and (2). We get,
\[\overrightarrow{PA} +\overrightarrow {PB}+ \overrightarrow{PC}^ + \overrightarrow {PD} = 2( \overrightarrow {PF} + \overrightarrow {PH} ) = 2 (2\overrightarrow {PQ} ) = 4 \overrightarrow{PQ} .\]