The side `A B` of a parallelogram `A B C D` is produced to any point `Pdot` A line through `A` parallel to `C P` meets `C B` produced in `Q` and the parallelogram `P B Q R` completed. Show that `a r(^(gm)A B C D)=a r(^(gm)B P R Q)dot` CONSTRUCTION : Join `A C` and PQ. TO PROVE : `a r(^(gm)A B C D)=a r(^(gm)B P R Q)`

Explanation

`ar(/_\ABC)=1/2" ar(parallelogram ABCD)"=>"eq(1)"`

`ar(/_\PBQ)=1/2ar"(parallelogram PBQR)"=>"eq(2)"`

`"In "/_\ACQ and /_\APQ`

`ar(/_\ACQ)=ar(/_\APQ)`

`ar(/_\ACQ)-ar(/_\ABQ)=ar(/_\APQ)-ar(ABQ)`

`ar(/_\ABC)=ar(/_\BQD)->"eq(3)"`

`therefore ar"(parallelogram ABCD)"=ar"(parallelogram PBQR)"`