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Triangles on the Same Base and between the Same Parallels

Diagonals AC ...

Question

Question 10
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

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Solution

It can be observed that $Δ D A C$ and $Δ D B C$ are on the same base DC and between same parallel AB and DC.
$∴ A r e a (Δ D A C) = A r e a (Δ D B C)$
Then, $A r e a (Δ D A C) - A r e a (Δ D O C) = A r e a (Δ D B C) - A r e a (Δ D O C)$
[Subtracting Area $(Δ D O C)$ from both sides]
$\Rightarrow A r e a (Δ A O D) = A r e a (Δ B O C)$

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Similar questions

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

A C

B D

A B C D

A B ∥ D C

O

(△ A O D) =

(△ B O C)

A C

B D

A B C D

A B ∥ D C

O

a r (A O D) = a r (B O C)

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