ABCD is a parallelogram in which diagonals AC and BD intersect at 0. If E, F, G and H are the mid points of AO, DO, CO and BO respectively, then the ratio of the perimeter of the quadrilateral EFGH to the perimeter of parallelogram ABCD is

We know that in a triangle the line segment joining the mid points of two sides, is parallel to the third side and measures half of the third side.
In the triangle OAB, EF is the line joining the mid points of OA and OB. Hence EF = $$\frac{1}{2}$$ AB.
In the triangle OBC, FG is the line joining the midpoints of OB and OC.
hence, FG = $$\frac{1}{2}$$ BC
similarly, GH = $$\frac{1}{2}$$ CD and HE = $$\frac{1}{2}$$ DA
perimeter of the quadrilateral EFGH = $$\frac{1}{2}$$ AB + $$\frac{1}{2}$$ BC + $$\frac{1}{2}$$ CD + $$\frac{1}{2}$$ DA
=$$\frac{1}{2}$$(Perimeter of parallelogram ABCD)
Hence,Option C is correct.