Question

𝐏 is a point in the interior of a parallelogram 𝐀𝐁𝐂𝐃. Show that:
(i) 𝐚𝐫(△𝐀𝐏𝐁) + 𝐚𝐫(△𝐏𝐂𝐃) = 𝟏/𝟐 𝐚𝐫(𝐀𝐁𝐂𝐃)
(ii) 𝐚𝐫(△𝐀𝐏𝐃) + 𝐚𝐫(△𝐏𝐁𝐂) = 𝐚𝐫(△𝐀𝐏𝐁) + 𝐚𝐫(△𝐏𝐂𝐃)

Answer 1

Assume two lines EF and GH such that EF ∥ AB and GH ∥ AD.
In ∥gm AEFB;
ar(APB) = ½ ​ar(AEFB)⟶(1)
In ∥g, EFCD;
ar(DPC) = ½ ar(EFCD)⟶(2)
From (1) and (2) (adding )
⇒ ar(APB) + ar(PDC) = ½ ​[ar(AEFB) + ar(EFCD)]
⇒ ar(APB) + ar(PDC)= ½ ar(ABCD)
In ∥gm AGHD;
ar(APD) = ½ ​ar(AGHD)⟶(3)
In ∥gm GBCH;
ar(PBC) = ½ ​ar(GBCH)⟶(4)
Adding (3) and (4)
⇒ ar(APD) + ar(PBC) = ½ [ar(AGHD) + ar(GBCH)]
⇒ ar(APD) + ar(PBC) = ½ ​ar(ABCD)
From previous part -
⇒ ar(APD) + ar(PBC) = ar(APB) + ar(PCD).
Hence, solved.