# EFGH is a parallelogram with diagonal EG and FH meeting at a point X. show that ar (traingle EXF) =ar(traingle GXF)=ar(triangle HXG)

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by mohit25

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by mohit25

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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.

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}