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) SR || AC and SR = $\dfrac{1}{2}AC$

(ii) PQ = SR

(iii) PQRS is a parallelogram

Given :

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

Mid-point theorem : 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.

(i) In △ ADC,

S and R are the mid-points of side AD and CD respectively.

By mid-point theorem,

⇒ SR || AC …….(1)

⇒ SR = $\dfrac{1}{2}AC$ ……(2)

Hence, proved that SR || AC and SR = $\dfrac{1}{2}AC$.

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

By using the mid-point theorem,

⇒ PQ || AC ………(3)

⇒ PQ = $\dfrac{1}{2}AC$ ……(4)

From equations (3) and (4), we get :

⇒ PQ = SR

Hence, proved that PQ = SR.

(iii) From equation (1) and (3), we get :

⇒ PQ || AC || SR

⇒ PQ || SR

Also,

⇒ PQ = SR (Proved above)

We know that,

If one pair of opposite sides are equal and parallel, then the figure is parallelogram.

Hence, proved that PQRS is a parallelogram.