Question

Abcd is a trapezium in which ab parallel to dc diagonals ac and bd intersect at o prove that ar(aod)=(∆boc)

Answer 1

Given: Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O.

To Prove: ar(Δ AOD) = ar(Δ BOC).

Proof: Δ ABD and Δ ABC are on the same base AB and between the same parallels AB and DC.
∴ ar(Δ ABD) = ar(Δ ABC) two triangles on the same base and between the same parallels are equal in area.


Subtract ar(Δ AOB) both sides,


⇒ ar(Δ ABD) - ar(Δ AOB) = ar(Δ ABC) - ar(Δ AOB)


⇒ ar(Δ AOD) = ar(Δ BOC).

Answer 2

let in ∆ABC and ∆DBC
the have the same base cd and between same parallel AB and CD
so area (ABC) = area (DBC)
subtracting ar(COD) from both sides
we get ar(aod) = (bod)
hope this helps u.
thanks