Question

Diagonals AC and BD of a trapezium ABCD with
AB || DC intersect each other at 0.
Prove that ar (AOD) = ar (BOC).​

Answer 1

Answer:

∆ABD = ∆ABC

( Triangles between two parallel lines and common base or both has same are as 1/2bh )

∆ABO + ∆AOD = ∆ABO + ∆ BOC

So

∆ AOD = ∆ BOC

Hence proved

Answer 2

Step-by-step explanation:

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