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Answer :

Let ABCD is a quadrilateral and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively.

\(\Rightarrow \) AS = SD, AP = BP, BQ = CQ and CR = DR.

We have to show that: PR and SQ bisect each other i.e., SO = OQ and PO = OR.

Now, in \(\triangle{ADC}\), S and R are mid-point of AD and CD.

We know that, the line segment joining the mid-points of two sides of a triangle is
parallel to the third side.

Thus, by midpoint theorem,

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

Similarly, in \(\triangle{ABC}\), P and Q are mid-point of AB and BC.

Thus, by midpoint theorem,

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

From Eq. (i) and (ii), we get,

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

Therefore, Quadrilateral PQRS is a parallelogram whose diagonals SQ are PR.

Also, we know that diagonals of a parallelogram bisect each other. So, and bisect each other.

Hence, it is proved that PR and SQ bisect each other.

- 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 rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.
- 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.
- 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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