In the given figure, ABCD is a trapezium in which AB || DC and diagonals AC and BD intersect at O. Prove that ar(ΔAOD) = ar(ΔBOC).

Question

Philoid · Accepted Answer

Here, in trapezium ABCD,

AB || DC and AC and BD are the diagonals intersecting at O.

Now, since ∆ACD and ∆BCD lie on the same base and between the same parallels.

Thus, ar(∆ACD) = ar(∆BCD)

Subtracting ar(∆COD) from both the sides, we get:

ar(∆ACD) − ar(∆COD) = ar(∆BCD)− ar(∆COD)

∴ ar(∆AOD) = ar(∆BOC)