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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.
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Solution
It is given that
Area (ΔAOD) = Area (ΔBOC)
Area (ΔAOD) + Area (ΔAOB) = Area (ΔBOC) + Area (ΔAOB)
Area (ΔADB) = Area (ΔACB)
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., AB || CD
Therefore, ABCD is a trapezium.
RELATED QUESTIONS
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
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ar (ABE) = ar (ACF)
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)
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In a ΔABC, if L and M are points on AB and AC respectively such that LM || BC. Prove
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(2) ar (ΔLBC) = ar (ΔMBC)
(3) ar (ΔABM) ar (ΔACL)
(4) ar (ΔLOB) ar (ΔMOC)
PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 cm2.
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In the following figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (APQ)
