Question.
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 $(\triangle A O D)=$ Area $(\triangle B O C)$
Area $(\triangle \mathrm{AOD})+$ Area $(\triangle \mathrm{AOB})=$ Area $(\triangle \mathrm{BOC})+$ Area $(\triangle \mathrm{AOB})$
Area $(\triangle \mathrm{ADB})=$ Area $(\triangle \mathrm{ACB})$
We know that triangles on the same base having areas equal to each other lie between the same parallels.
Therefore, these triangles, ΔADB and ΔACB, are lying between the same parallels.
i.e., $A B \| C D$
Therefore, ABCD is a trapezium.
It is given that
Area $(\triangle A O D)=$ Area $(\triangle B O C)$
Area $(\triangle \mathrm{AOD})+$ Area $(\triangle \mathrm{AOB})=$ Area $(\triangle \mathrm{BOC})+$ Area $(\triangle \mathrm{AOB})$
Area $(\triangle \mathrm{ADB})=$ Area $(\triangle \mathrm{ACB})$
We know that triangles on the same base having areas equal to each other lie between the same parallels.
Therefore, these triangles, ΔADB and ΔACB, are lying between the same parallels.
i.e., $A B \| C D$
Therefore, ABCD is a trapezium.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.