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Question
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that arAOD = arBOC. Prove that ABCD is a trapezium.
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Solution
It is given that AOD = BOC . Now, add DOC in both triangles
This will give ACD= BCD.
Now there is a theorem that parallelograms on the same base and between the same parallels have equal area.
Now in this condition , The areas of both triangles are equal and they are also on the same base DC so they should be between the same parallels according to the theorem.
THis way AB is parallel to DC.One pair of opposite sides is parallel .
Hence it is proved that ABCD is a trapezium
It is given that AOD = BOC . Now, add DOC in both triangles
This will give ACD= BCD.
Now there is a theorem that parallelograms on the same base and between the same parallels have equal area.
Now in this condition , The areas of both triangles are equal and they are also on the same base DC so they should be between the same parallels according to the theorem.
THis way AB is parallel to DC.One pair of opposite sides is parallel .
Hence it is proved that ABCD is a trapezium
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