In Figure, `P` is a point in the interior of a parallelogram `A B C Ddot` Show that `a r( A P B)+a r( P C D)=1/2a r\ (|""|^(gm)A B C D)` `a r\ ( A P D)+a r\ ( P B C)=a r\ ( A P B)+a r( P C D)`

Explanation

Given ABCD is a parallelogram. So, `AB||CD, AD||BC`.
(i) Now, draw MPR parallel to AB and CD both and also draw a perpendicular PS on AB.
`.: MR||AB and AM||BR`
`:. ABRM` is a parallelogram
`:. Ar(||gm ABRM) = AB xx PS`...(1)
and `ar(DeltaAPB) = (1)/(2) xx AB xx PS`
`rArr ar(DeltaAPB) = (1)/(2) ar(||gm ABRM)` [from (1)]
Similarly, `ar(DeltaPCD) = (1)/(2) ar(||gm MRCD)`
Now, `ar(DeltaAPB) + ar(DeltaPCD) = (1)/(2) ar(||gm ABRM) + (1)/(2) ar (||gm MRCD)`
`=(1)/(2) ar(||gm ABCD)`...(2)
(ii) Similarly, we can draw a line through P parallel to AD and through the point P draw perpendicular on AD, we can prove that
`ar(DeltaAPD) + ar(DeltaPBC) = (1)/(2) ar(||gm ABCD)`
From (2) and (3), we get
`ar(DeltaAPD) + ar(DeltaPBC) = ar(DeltaAPB) + ar(DeltaPCD)`