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# ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show thati) SR || AC and SR = (1/2) AC ii) PQ = SRiii) PQRS is a parallelogram.

Given: P, Q, R, and S are mid-points of the sides.
Therefore, AP = PB, BQ = CQ CR = DR and AS = DS

i) Now, in $$\triangle{ADC}$$, we have,
S is the midpoint of AD and R is the mid point of CD.
As, we know that,
By midpoint theorem, the line segment joining the mid points of two sides of a triangle is parallel to the third side.
Thus, we can say that, SR || AC ...(i)
and SR = ($$\frac{1}{2}$$ ) AC ...(ii)

ii) Similarly, now, in $$\triangle{ABC}$$, we have,
PQ || AC ...(iii)
and PQ = ($$\frac{1}{2}$$ ) AC ...(iv)
Now, from (ii) and (iv), we get,
SR = PQ = ($$\frac{1}{2}$$ ) AC ...(v)

Now, from (i), (iii) and (v), we get,
PQ || SR and PQ = SR.
$$\therefore$$ PQRS is a parallelogram
(Since, a pair of opposite sides of a quadrilateral PQRS is equal and parallel)
Hence, proved.