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:
It is given that
Area $(\triangle \mathrm{AOD})=$ Area $(\Delta \mathrm{BOC})$
Area $(\triangle \mathrm{AOD})+$ Area $(\triangle \mathrm{AOB})=$ Area $(\triangle \mathrm{BOC})+$ Area $(\triangle \mathrm{AOB})$
Area $(\triangle A D B)=$ Area $(\triangle A C B)$
We know that triangles on the same base having areas equal to each other lie between the same parallels.
Therefore, these triangles, $\triangle \mathrm{ADB}$ and $\triangle \mathrm{ACB}$, are lying between the same parallels.
i.e., $A B \| C D$
Therefore, $A B C D$ is a trapezium.
![](/media/uploads/2021/12/Diagonals-AC-and-BD-of-a-quadrilateral-ABCD-intersect-at-O-in-such-a-way-that-ar.png)
It is given that
Area $(\triangle \mathrm{AOD})=$ Area $(\Delta \mathrm{BOC})$
Area $(\triangle \mathrm{AOD})+$ Area $(\triangle \mathrm{AOB})=$ Area $(\triangle \mathrm{BOC})+$ Area $(\triangle \mathrm{AOB})$
Area $(\triangle A D B)=$ Area $(\triangle A C B)$
We know that triangles on the same base having areas equal to each other lie between the same parallels.
Therefore, these triangles, $\triangle \mathrm{ADB}$ and $\triangle \mathrm{ACB}$, are lying between the same parallels.
i.e., $A B \| C D$
Therefore, $A B C D$ is a trapezium.
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