Diagonals AC , BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) =ar (BOC).

Asked by  | 13th Sep, 2008, 04:12: PM

Expert Answer:

Denote

the Area (triangle AOD) as Area(I)

The Area (triangle BOC ) as Area(II)

Area (triangle AOB) as Area(III)

Given that

Area(I)=Area(II)

So,

Area(I)+Area(III)=Area(II)+Area(III)

This gives us that,

Area(triangle ADB)=Area(triangle ACB)

So, AB must be parallel to DC ..(as triangles on the same base and equal in area must lie between same parallels)

So, ABCD is a trapezium as one pair of opposite sides is parallel.

 

Answered by  | 22nd Jan, 2009, 01:45: PM

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