The side `A B` of a parallelogram `A B C D` is produced to any point `Pdot` A line through `A` and parallel to `C P` meets `C B` produced at `Q` and then parallelogram `P B Q R` is completed as shown in Figure. Show that `a r(|"|"hatgm\ A B C D)=a r\ (|""|^(gm)P B Q R)`

Explanation

Join AC, CP, PQ and QA.
Now, `triangleAQC and triangleAQP` being on the same base AQ and between the same parallels AQ and CP, we have
`ar(triangleAQC)=ar(triangleAQP)`
`rArr ar(triangleAQC
)-ar(triangleAQB)=ar(triangleAQP)-ar(triangleAQB)`
`rArr ar(triangleABC)=ar(triangleAQB)=ar(triangleAQP)-ar(AQB)`
`rArr (1)/(2)ar("||gm ABCD")=(1)/(2)("||gm BQRP")`
[`therefore` AC divides ||gm ABCD into two `triangle` of equal area and QP divides ||gm BQRP into two `triangle` of equal area]
`rArr ar("||gm ABCD")=ar("||gm BQRP).`
Hence, `ar("||gm ABCD")=ar("||gm BQRP")`.