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 $\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}+\overrightarrow{PD}=4\overrightarrow{PQ}$, 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, G and H will be a parallelogram. Hence its diagonals will bisect each other, say at Q.
Now, F is the midpoint of BC.
$$\begin{gathered} \frac{\stackrel{\rightharpoonup }{PB}+\stackrel{\rightharpoonup }{PC}}{2}=\stackrel{\rightharpoonup }{PF} \\ \therefore \stackrel{\rightharpoonup }{PB}+\stackrel{\rightharpoonup }{PC}=2\stackrel{\rightharpoonup }{PF}.....\left(1\right) \end{gathered}$$
And, H is the midpoint of AD.
$$\begin{gathered} \frac{\stackrel{\rightharpoonup }{PA}+\stackrel{\rightharpoonup }{PD}}{2}=\stackrel{\rightharpoonup }{PH} \\ \therefore \stackrel{\rightharpoonup }{PA}+\stackrel{\rightharpoonup }{PD}=2\stackrel{\rightharpoonup }{PH}.....\left(2\right) \end{gathered}$$
Adding (1) and (2). We get,
$\stackrel{\rightharpoonup }{PA}+\stackrel{\rightharpoonup }{PB}+\stackrel{\rightharpoonup }{PC}+\stackrel{\rightharpoonup }{PD}=2(\stackrel{\rightharpoonup }{PF}+\stackrel{\rightharpoonup }{PH})=2\left(2\stackrel{\rightharpoonup }{PQ}\right)=4\stackrel{\rightharpoonup }{PQ}.$