If `A B C D` is a parallelogram, the prove that `a r( A B D)=a r( B C D)=a r( A B C)=a r( A C D)`=1/2a r`(||^(gm)`A B C D)

Explanation

Given:`ABCD` is a parallelogram.
When we join the diagonal of parallelogram, it divides it into two quadrilaterals.
Proof:
Let `AC` is the diagonal, then
`ar(ΔABC)=ar(ΔACD)=1/2ar(ABCD)`
Let `BD` be another diagonal
`ar(ΔABD)=ar(ΔBCD)=1/2ar(ABCD)`
Now, we have
`ar(ΔABC)=ar(ΔACD)`
`=ar(ΔABD)=ar(ΔBCD)`
`=1/2ar(ABCD)`
Hence Proved.