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

Expert Answer:

Consider the following diagram.
 
 
Now consider the triangles DOP and BOQ.
 
 
angle D O P equals angle B O Q space space space space open square brackets v e r t i c a l l y space o p p o s i t e space a n g l e s space a r e space e q u a l close square brackets
OD = OB [diagonals of a parallelogram bisect at O]
 
angle O D P equals angle O B Q space open square brackets A l t e r n a t e space i n t e r i o r space a n g l e s close square brackets
Thus, by Angle-Side-Angle criterion of congruence, triangle D O P approximately equal to triangle B O Q.
 
Since congruent triangles have equal areas, we have
 
a r open parentheses triangle D O P close parentheses equals a r open parentheses triangle B O Q close parentheses...(1)
The diagonal BD divides the parallelogram into two triangles, triangle B C D space a n d space triangle A B D of equal areas.
 
Therefore, we have
 
a r open parentheses triangle A B D close parentheses equals a r open parentheses triangle B C D close parentheses...(2)
Now consider the quadrilateral QCDP.
 
a r open parentheses Q C D P close parentheses equals a r open parentheses triangle B C D close parentheses plus a r open parentheses triangle D O P close parentheses minus a r open parentheses triangle B O Q close parentheses space space space space open square brackets f r o m space left parenthesis 1 right parenthesis space a n d space left parenthesis 2 right parenthesis close square brackets rightwards double arrow a r open parentheses Q C D P close parentheses equals a r open parentheses A P Q B close parentheses
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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