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i) SR || AC and SR = (1/2) AC

ii) PQ = SR

iii) PQRS is a parallelogram.

Answer :

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.

- 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.
- 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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