Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
Solution:
We have quadrilateral ABCD. Diagonals AC and BD intersect each other at O.
It is given that
(1)
We can see that
and
are on the same base CD. (2)
From (1) and (2), we can say that
(Two triangles having the same base and equal areas lie between the same parallels.)
ABCD is a trapezium.
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