if p is a point in interior of parallelogram abcd prove that at apd < 1/2 ar parallelogram abcd
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The diagonal of a //gm divides it into 2 congruent triangles which are equal in area and the area of each of those triangles is half are of the //gm.
Here in //gm ABCD,
ABC = ADC = 1/2 ABCD.
Therefore, for APD to be equal to the 1/2 ar ABCD the point P should lie somewhere on BC.
As this is contradictory to the statement in the question, point p lies in interior of ABCD, APD is not equal to 1/2 ar ABCD.
No triangle in a //gm can be higher in area than half area of the //gm.
Therefore, APD < 1/2 //gm ABCD
Here in //gm ABCD,
ABC = ADC = 1/2 ABCD.
Therefore, for APD to be equal to the 1/2 ar ABCD the point P should lie somewhere on BC.
As this is contradictory to the statement in the question, point p lies in interior of ABCD, APD is not equal to 1/2 ar ABCD.
No triangle in a //gm can be higher in area than half area of the //gm.
Therefore, APD < 1/2 //gm ABCD
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