Question 15
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.

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Solution

It is given that,
$A r e a (Δ A O D) = A r e a (Δ B O C)$
$A r e a (Δ A O D) + A r e a (Δ A O B) = A r e a (Δ B O C) + A r e a (Δ A O B)$ [Adding Area $(Δ A O B)$ to both sides]
$A r e a (Δ A D B) = A r e a (Δ A C B)$
We know that triangles on the same base having areas equal to each other lie
between the same parallels.
Therefore, these triangles, $Δ A D B$ and $Δ A C B$ , are lying between the same parallels.
i.e., AB || CD
Therefore, ABCD is a trapezium.

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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.

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