Question

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that:
(i)S R || A C and S R=$\dfrac{1}{2}$A C
(ii) P Q =S R
(iii) PQRS is a parallelogram

Answer 1

Hint – In order to solve this problem use the mid-point theorem in triangles and also use the definition of parallelogram. Doing this you will get the right answer.

Complete step-by-step answer:
(i)We need to show that SR || AC and SR = $\dfrac{1}{2}$AC (Given).
On considering triangle ADC,
S and R are the mid points of two sides of triangle AD and DC.
As we know that the Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.
So, from the above statement clearly we can say,
SR || AC and SR = $\dfrac{1}{2}$AC……………..(1)
(ii) Now we have to show that PQ = SR.
As P and Q are the midpoints of the sides AB and BC of triangle ABC (Given).
On considering triangle ABC and applying the mid-point theorem in it we can say that PQ || AC and PQ = $\dfrac{1}{2}$AC……..(2)
So, from (1) and (2) we can say PQ || SR and also PQ = SR………(3)
(iii) We also need to show that PQRS is a parallelogram.
 So, as we know the quadrilateral in which two of the sides are equal and parallel is a parallelogram.
We see from (3) that PQ || SR and PQ = SR therefore from the definition of parallelogram we can say PQRS is a parallelogram.

Note – To solve this problem you should know the theorem of midpoint of triangle and definition of parallelogram. Part (i), (ii), (iii) are linked with each other. Squares, Rectangles and Rhombuses are all Parallelograms and the mid-point theorem is valid in all the triangles.