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Question

The side AB of a parallelogram ABCD is produced to any point P.
A line through A a
nd parallel to CP meets CB produced at Q and then parallelogram PBQR i completed   (see the following figure).
Show that ar (ABCD)=ar(PBQR).  


 


Solution



Let us join AC and PQ.
ΔACQ and ΔAQP are on the same base AQ and between the same parallels AQ and CP.
ar(ΔACQ)=ar(ΔAPQ)

ar(ΔACQ)ar(ΔABQ)=ar(ΔAPQ)ar(ΔABQ)
[Subtracting ar (ΔABQ) from both sides]
ar(ΔABC)=ar(ΔQBP).......(1)

Since AC and PQ are diagonals of parallelograms ABCD and PBQR respectively,
ar(ΔABC)=12ar(ABCD)...(2)

ar(ΔQBP)=12ar(PBQR)...(3)

From equations (1), (2), and (3), we obtain
12ar(ABCD)=12ar(PBQR)

ar(ABCD)=ar(PBQR)

 

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