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Answer :

Given: ABCD is a rhombus and P, Q, R and S are mid-points of AB, BC, CD and DA.

By midpoint theorm, Now, in \(\triangle{ADC}\), we have,

Thus, SR || AC ...(i)

and SR = (\(\frac{1}{2} \) ) AC ...(ii)

Similarly, in \(\triangle{ABC}\), we have,

PQ || AC ...(iii)

and PQ = (\(\frac{1}{2} \) ) AC ...(iv)

from (i), (ii), (iii) and (iv), we get,

SR = PQ = (\(\frac{1}{2} \) ) AC and PQ || SR

\(\therefore \) Quadrilateral PQRS is a parallelogram.

Now, we know that diagonals of a rhombus bisect each other at right angles.

\(\therefore \) \(\angle{EOF}\) = \(90^\circ\).

Now, By midpoint theorem, we have,

RQ || BD

Thus, RE || OF

As SR || AC ...(from (i))

Thus, FR || OE

\(\therefore \) OERF is a parallelogram

So, \(\angle{ERF}\) = \(\angle{EOF} = 90° \)

(since, Opposite angle of a quadrilateral is equal)

Thus, PQRS is a parallelogram with \(\angle{R}= 90° \)

Hence, it is proved that PQRS is a rectangle.

- 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.
- ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rhombus.
- ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid-point of BC.
- In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see figure). Show that the line segments AF and EC trisect the diagonal BD.
- Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
- ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that i) D is the mid-point of AC ii) MD is perpendicular to AC iii) CM = MA = (\(\frac{1}{2} \)) AB

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