Question

The side AB of a parallelogram ABCD is produced to any point P. A line through A, parallel to CP meets CB produced at Q and the parallelogram PBQR is completed. Prove that ar (||gm ABCD) = ar (||gm PBQR).

Answer 1

Given: The side AB of a parallelogram ABCD is produced to P. 

AQ || CP where Q is on CB produced. PBQR is a parallelogram.

To prove: ar(||gm ABCD) = ar(||gm PBQR)

Construction: Join AC and QP

Proof: In parallelogram ABCD, AC is a diagonal

ar (||gm ABCD) = 2 ar (∆ABC) …(i)

Similarly, in parallelogram PBQR, QP is a diagonal

ar (||gm PBQR) = 2 ar (∆BPQ) …(ii)

Now ∆AQC and ∆AQP are on the same base AQ and between the same parallels AQ and CP.

ar (∆AQC) = ar (∆AQP)

ar (∆AQC) – ar (AAQB) = ar (∆AQP) – ar (∆AQB)

⇒ ar (∆ABC) = ar (∆BPQ) …(iii)

From (i), (ii) and (iii), we get

ar (||gm ABCD) = ar (||gm PBQR).