Given : PQRS and ABRS are parallelograms and X is any point on side BR.

To prove : (i) ar (PQRS) = ar (ABRS) (ii) ar (AXS) = 1 2 ar (PQRS)

Proof : (i) In ∆ASP and BRQ, we have

∠SPA = ∠RQB [Corresponding angles] ...(1)

∠PAS = ∠QBR [Corresponding angles] ...(2)

∴ ∠PSA = ∠QRB [Angle sum property of a triangle] ...(3)

Also, PS = QR [Opposite sides of the parallelogram PQRS] ...(4)

So, ∆ASP ≅ ∆BRQ [ASA axiom, using (1), (3) and (4)]

Therefore, area of ∆PSA = area of ∆QRB [Congruent figures have equal areas] ...(5)

Now, ar (PQRS) = ar (PSA) + ar (ASRQ]

= ar (QRB) + ar (ASRQ]

= ar (ABRS)

So, ar (PQRS) = ar (ABRS) **Proved**.

(ii) Now, ∆AXS and ||gm ABRS are on the same base AS and between same parallels AS and BR

∴ area of ∆AXS = 1/ 2 area of ABRS

⇒ area of ∆AXS = 1/ 2 area of PQRS [ ar (PQRS) = ar (ABRS]

⇒ ar of (AXS) = 1/ 2 ar of (PQRS) **Proved.**