Question

Diagonals AC and BD of a trapezium ABCD with AB||DC intersect each other at O.Prove that ar(AOD)=ar(BOC).

Answer 1

Solution:

Given that AB║DC

To prove ar(ΔAOD) =ar(ΔBOC)

Proof: In ΔABC and ΔABD

They are between same ║⇒AB║DC

Now, ar(ΔABC) = ar(ΔABD) -----(1)

Subtracting ΔAOB from (1) we get

ar(ΔABC) - ar(AOB) = ar(ABD) - ar(AOB)

ar(ΔBOC) = ar(ΔAOD) Hence proved

Answer 2

Dear user:

Given:

AB||DC

TO PROVE:

ar(AOD) =ar(BOC)

Proof:

ΔADC and ΔBDC are on the same base DC and between same parallel AB and DC.

∴ ar(ΔADC) = ar(ΔBDC) [triangles on the same base and between same parallel are equal in area]

Subtract Area (ΔDOC) from both side.

ar(ΔADC) – ar(ΔDOC) = ar(ΔBDC) – ar(ΔDOC)

ar(ΔAOD) = ar(ΔBOC)

Hence proved☺☺

#AN