The diagonals of a parallelogram ABCD intersect at a 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 equal parts of equal area.
Asked by Sthitaprajna Mishra | 4th Feb, 2014, 08:17: PM
Consider the following diagram.
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
The diagonal BD divides the parallelogram into two triangles, of equal areas.
Therefore, we have
Now consider the quadrilateral QCDP.
Thus PQ divides the parallelogram ABCD into two equal quadrialterals QCDP and AQCB.
Answered by | 5th Feb, 2014, 11:34: AM
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