Question

P is a point in the interior of a parallelogram ABCD show that ar (triangle APB)+ar (triangle PCB)=1/2 ar (ABCD) ;ar(triangle APD)+ar(triangle PBC=ar(triangle APB)+ar(triangle PCD).(Hint:through P,draw a line parallel to AB).

Answer 1

this is the answer of this question please tell me will u be able to understand the answer or not

Answer 2

Answer:

Step-by-step explanation: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

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