NCERT Solutions
Class 9 Maths
Areas of Parallelograms and Triangles
Ex.9.2 Q.4
In the figure, P is a point in the interior of a
parallelogram ABCD. Show that
(i) ar (APB) + ar (PCD) = 
(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)
Given: A parallelogram ABCD. P is a point inside it.
To prove:
(i) ar (APB) + ar (PCD) = 
(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)
Construction: Draw EF through P parallel to AB, and GH through P parallel to AD.
Proof: In parallelogram FPGA, AP is a diagonal,
So, area of ∆APG = area of ∆APF ..............(1)
In parallelogram BGPE, PB is a diagonal,
So, area of ∆BPG = area of ∆EPB ..............(2)
In parallelogram DHPF, DP is a diagonal,
area of ∆DPH = area of ∆DPF ..........(3)
In parallelogram HCEP, CP is a diagonal,
So, area of ∆CPH = area of ∆CPE ......(4)
Adding equation 1, 2, 3 and 4, we get
area of ∆APG + area of ∆BPG + area of ∆DPH + area of ∆CPH
= area of ∆APF + area of ∆EPB + area of ∆DPF + area ∆CPE
=> [area of ∆APG + area of ∆BPG] + [area of ∆DPH + area of ∆CPH]
= [area of ∆APF + area of ∆DPF] + [area of ∆EPB + area of ∆CPE]
=> area of ∆APB + area of ∆CPD = area of ∆APD + area of ∆BPC ……... (5)
But area of parallelogram ABCD
= area of ∆APB + area of ∆CPD + area of ∆APD + area of ∆BPC ………. (6)
From equation 5 and 6, we get
area of ∆APB + area of ∆PCD = area of 
=> ar (APB) + ar (PCD) =
Hence, Proved.
From equation 5,
=> ar (APD) + ar (PBC) = ar (APB) + ar (CPD)