The diagonals of a parallelogram ABCD intersect at point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area that is ar (APQB) = ar (PQCD).



Asked by Topperlearning User | 17th Aug, 2017, 01:53: PM

Expert Answer:

Diagonals of a parallelogram divide it into two triangles of equal area.
ar (ABC) = ar(ACD)
ar (ABQO) + ar (COQ) = ar (CDPO) + ar (AOP)                          ...(i)
AOP COQ                               (ASA)
ar (AOP) = ar (COQ)… (ii)
From (i) and (ii)
 ar (ABQO) + ar (COQ) = ar (CDPO) + ar (AOP)
 ar (ABQP) = ar (PQCD)

Answered by  | 17th Aug, 2017, 03:53: PM