Show that the lines joining the mid points of the consecutive sides of a quadliteral form a parallelogram. Using vector method -
Show that the lines joining the mid points of the consecutive sides of a quadliteral form a parallelogram.
The link from which i am taking hint suggest to take one point as origin. But i am unable to understand.
As this question involves use of vectors so problem in taking mid points.
Can someone please explain actual procedure to solve these type of questions.
And i have spent so much time to find solution but no link involve use of vectors. So its become more difficult.
 A: We may use the same Lemma (and the same notation) that was crucial in your other question:

If $A,B,C,D$ are four points in the plane, $ABCD$ is a parallelogram iff $A+C=B+D$.

In the present case, we have four generic points in the plane, $P,Q,R,S$. Then the midpoints of the $PQ,QR,RS,SP$ segments are given, respectively, by:
$$ \frac{P+Q}{2},\quad\frac{Q+R}{2},\quad \frac{R+S}{2},\quad\frac{S+P}{2} $$
and is is straightforward to check that
$$ \frac{P+Q}{2}+\frac{R+S}{2} = \frac{Q+R}{2}+\frac{S+P}{2} $$
holds since both terms equal $\frac{P+Q+R+S}{2}$. As a consequence:

If $A,B,C,D$ are four distinct points in the plane, the midpoints of
  the segments $AB,BC,CD,DA$ are the vertices of a parallelogram.

Namely, Varignon's parallelogram, whose sides are parallel to the diagonals of $ABCD$.
Remarkably, the area of the Varignon's parallelogram is just half the area of $ABCD$:

A: This property is true in any real vector space E ($\Bbb R^2, \Bbb R^3,..., \Bbb R^{666},... $ for example to be simple).
We recall, since we are asked to use vector methods, that a vector is neither more nor less than an element of a vector space. We can limit ourselves to $\Bbb R^3$ if we wish, but it would be a shame to limit ourselves to $\Bbb R^2$.
Let $a,b,c,d$ four points in $E$.
By definition, the midpoint of $ab$ is $x=a+\frac12(b-a)=\frac12(a+b)$; let $y,z,t$ those from $bc, cd, da$.
Then, $y-x=\frac12(b+c)-\frac12(a+b)=\frac12(c-a)$.
And $z-t=\frac12(c+d)-\frac12(a+d)=\frac12(c-a)$.
Then $y-x=z-t$.
Then, by definition, $(x,y,z,t)$ is said to be a parallelogram.
