Question

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).

Answer 1

Given : A parallelogram ABCD. P and Q are any points on DC and AD respectively. 

To prove : ar (APB) = ar (BQC) 

Construction : Draw PS || AD and QR || AB. 

Proof : In parallelogram ABRQ, BQ is the diagonal. 

∴ area of ∆BQR = 1 /2 area of ABRQ ... (i)

In parallelogram CDQR, CQ is a diagonal. 

∴ area of ∆RQC = 1 /2 area of CDQR ... (ii) 

Adding (i) and (ii), we have 

area of ∆BQR + area of ∆RQC 

= 1/ 2 [area of ABRQ + area of CDQR] 

⇒ area of ∆BQC = 1/ 2 area of ABCD ... (iii) 

Again, in parallelogram DPSA, AP is a diagonal. 

∴ area of ∆ASP = 1/ 2 area of DPSA ... (iv) 

In parallelogram BCPS, PB is a diagonal. 

∴ area of ∆BPS = 1/ 2 area of BCPS ... (v) 

Adding (iv) and (v) 

area of ∆ASP + area of ∆BPS = 1/ 2 (area of DPSA + area of BCPS) 

⇒ area of ∆APB = 1 2 (area of ABCD) ... (vi) 

From (iii) and (vi), we have 

area of ∆APB = area of ∆BQC. Proved.