Math, asked by deepak70, 1 year ago

diagonals of parallelograms ABCD intersect at point O.Through O a line drawn to intersect AD at P and BC at Q.show that PQ divides the parallelogram into two parts of equal area.

Answers

Answered by edwinthomas060
24

  Now consider the triangles DOP and BOQ.  OD = OB [diagonals of a parallelogram bisect at O] Thus, by Angle-Side-Angle criterion of congruence, . Since congruent triangles have equal areas, we have ...(1)The diagonal BD divides the parallelogram into two triangles,  of equal areas. Therefore, we have ...(2)Now consider the quadrilateral QCDP. Thus PQ divides the parallelogram ABCD into two equal quadrialterals QCDP and AQCB.

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