In a trapezoid abcd with legs ab and cd, the diagonals intersect each other at point o. compare the areas of △abo and △cdo.

ar(ΔABO) = ar(ΔCDO)

Explanation:

The image attached below.

Given ABCD is a trapezoid with legs AB and CD.

AB and CD are non-parallel sides between the parallels AD and BC.

In ΔABD and ΔACD,

We know that, triangles lie between the same base and same parallels are equal in area.

⇒ AD is the common base for ΔABD and ΔACD and they are lie between the same parallels AD and BC.

Hence, ar(ΔABD) = ar(ΔACD) – – – – (1)

Now consider ΔABO and ΔCDO,

Subtract ar(ΔAOD) on both sides of (1), we get

ar(ΔABD) – ar(ΔAOD) = ar(ΔACD) – ar(ΔAOD)

⇒ar(ΔABO) = ar(ΔCDO)

Hence, ar(ΔABO) = ar(ΔCDO).