Given : A parallelogram ABCD. P is a point inside it.
To prove : (i) ar (APB) + ar(PCD)
= 1/ 2 ar (ABCD)
(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,
∴ area of ∆APG = area of ∆APF ... (i)
In parallelogram BGPE, PB is a diagonal,
∴ area of ∆BPG = area of ∆EPB ... (ii)
In parallelogram DHPF, DP is a diagonal,
∴ area of ∆DPH = area of ∆DPF ... (iii)
In parallelogram HCEP, CP is a diagonal,
∴ area of ∆CPH = area of ∆CPE ... (iv)
Adding (i), (ii), (iii) and (iv)
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 ... (v)
But area of parallelogram ABCD
= area of ∆APB + area of ∆CPD + area of ∆APD + area of ∆BPC ... (vi)
From (v) and (vi)
area of ∆APB + area of ∆PCD = 1/ 2 area of ABCD
or, ar (APB) + ar (PCD) = 1/ 2 ar (ABCD) Proved.
(ii) From (v),
⇒ ar (APD) + ar (PBC) = ar (APB) + ar (CPD) Proved
