Answers

2016-04-17T11:53:02+05:30
Given,
EFGH is a parallelogram with diagonals EG and FH meeting at a point X.

In parallelogram EFGH, we have
EG = FH (Diagonals of a llgm are equal)
EF = HG (Opposite sides of llgm are equal)
FG = FG (Common)

Therefore, triangle EFG is congruent to triangle HGF by SSS congruence criteria.

That implies that ar(triangle EFG) = ar(triangle HGF)
ar(triangle EFG) = ar(triangle EXF) + ar(triangle GXF)               ......(1)
ar(triangle HGF) = ar(triangle HXG) + ar(triangle GXF)              ......(2)

From (1) and (2), we have
ar(triangle EXF) + ar(triangle GXF) = ar(triangle HXG) + ar(GXF)
ar(triangle EXF) = ar(triangle GXF) = ar(triangle HXG)
Hence, proved.
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2016-04-17T11:54:43+05:30
X is the mid point of both the diagonals[diagonals of a parallelogram bisect each other]
In tri. EFG
XF is the median
therefore ar (EXF)=ar (GXF)[since, median of a triangle divides the triangle into two triangles of equal area]
In tri. FGH
GX is the median
therefore ar (HXG)=ar (GXF) [since, median of a triangle divides the triangle into two triangles of equal area]

therefore ar (EXF)=ar (GXF)=ar (HXG)

{Hence proved}
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