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).
