Question

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.20). AC is a diagonal. Show that : 

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

(ii) PQ = SR 

(iii) PQRS is a parallelogram.

Answer 1

(i) In ∆ADC, S and R are the mid-points of sides AD and CD respectively. 

In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it. 

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

(ii) In ∆ABC, P and Q are mid-points of sides AB and BC respectively. 

Therefore, by using mid-point theorem, 

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

Using equations (1) and (2), we obtain 

PQ || SR and PQ = SR ... (3) 

∠PQ = SR

(iii) From equation (3), we obtained 

PQ || SR and PQ = SR 

Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal.

Hence, PQRS is a parallelogram.