The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed as shown in Figure. Show that `ar(|\|^(gm) ABCD)=ar (|\|^(gm) PBQR)`

Explanation

Solution
Construction
Let us join `AC` and `PQ`.

`/_\ACQ` and `/_\AQP` are on the same base `AQ` and are between same parallels `AQ` and `CP.`
`:. Area (/_\ACQ)= Area(/_\APQ)`
`=>Area(/_\ACQ)`−Area`(/_\ABQ)`=Area`(/_\ACQ)`−Area`(/_\ABQ)`
`=>`Area `(/_\ABC)`= Area `(/_\ABQ) ... (1)`
Since `AC` and `PQ` are diagonals of parallelograms `ABCD` and `PBQR` respectively,
So,
Area `(/_\ABC)= 1/2` Area of parallelogram `ABCD ... (2)`
Area `(/_\QBP)= 1/2` Area of parallelogram `BPRQ ... (3)`
From equations `(1),(2)`, and `(3)`, we get
`1/2`Area of parallelogram `ABCD= 1/2` Area of parallelogram `BPRQ`
`:.`Area of parallelogram `ABCD=` Area of parallelogram `BPRQ`
Hence proved, areas of both parallelograms are equal.