In the figure, given below, 2AD = AB, P is mid-point of AB, Q is mid-point of DR and PR // BS. Prove that :

(i) AQ // BS

(ii) DS = 3RS

Given,

P is mid-point of AB.

∴ AP = PB

Since, 2AD = AB

∴ AD = $\dfrac{AB}{2}$

∴ AP = AB = AD.

(i) By mid-point theorem,

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

In △ DPR,

A and Q are mid-points of DP and DR respectively.

∴ AQ || PR [By mid-point theorem] …….(1)

Given,

⇒ PR || BS ………..(2)

From equations (1) and (2), we get :

⇒ AQ || BS.

Hence, proved that AQ || BS.

(ii) By equal intercept theorem,

If a transversal makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts.

From figure,

PR || BS

Since, AD = AP = PB

∴ DQ = QR = RS ………(3)

From figure,

⇒ DS = DQ + QR + RS

⇒ DS = RS + RS + RS [From equation (3)]

⇒ DS = 3RS.

Hence, proved that DS = 3RS.