Question

P is a point in the interior of a
parallelogram ABCD. Show that
(1) ar(APB)-ar(PCD) = ar (ABCD)
2
(i) ar (APD)-ar (PBC)=ar (APB)+ar (PCD)
(Hint: Through P. draw a line parallel to AB.]
​

Answer 1

Answer:

Assume two lines EF and GH such that EF∥AB and GH∥AD.

In ∥gmAEFB;ar(APB)=

2

1

ar(AEFB)⟶(1)

In ∥g,EFCD;ar(DPC)=

2

1

ar(EFCD)⟶(2)

From (1) and (2) (adding )

⇒ar(APB)+ar(PDC)=

2

1

[ar(AEFB)+ar(EFCD)]

⇒ar(APB)+ar(PDC)=

2

1

ar(ABCD)−

In ∥gmAGHD;ar(APD)=

2

1

ar(AGHD)⟶(3)

In ∥gmGBCH;ar(PBC)=

2

1

ar(GBCH)⟶(4)

Adding (3) and (4)

⇒ar(APD)+ar(PBC)=

2

1

[ar(AGHD)+ar(GBCH)]

⇒ar(APD)+ar(PBC)=

2

1

ar(ABCD)

From previous part -

⇒ar(APD)+ar(PBC)=ar(APB)+ar(PCD).

Hence, solved.