If the mid-points of the sides of a 4-gon(also known as a quadrilateral, but we prefer to call it '4-gon') are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon.

Let \(ABCD\) be a quadrilateral.
Let \(P, Q, R, S\) be the midpoints of \(AB, BC,O CD, \) and \(DA\) respectively. Join \(PQ, QR, RS,O SP\).
In triangle \(ABC\), since \(P\) and \(Q\) are midpoints,
\(PQ || AC\)
Similarly, in triangle \(ADC\),
\(SR || AC\)
Hence, \(PQ || SR\) Similarly, in triangle \(BCD\) and \(BAD\),
\(QR || BD\) and
\(PS || BD\)
Hence, \(QR || PS\)
Since both pairs of opposite sides are parallel,
\(PQRS\) is a parallelogram.

Using properties of midpoint triangles, the four corner triangles occupy exactly half the area of the original quadrilateral.
The remaining central parallelogram occupies the other half.
Therefore,
\(Area\,of \,parallelogram \,PQRS\, = \frac{1}{2}\,Area \,of \,quadrilateral\,ABCD\)
Hence proved.