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# Diagonals AC and BD of a trapezium ABCD with AB ll DC intersect each other at O. Prove that Area of Δ (AOD) = Area of Δ (COB) and Δ AOB and Δ COD are similar.

Last updated date: 15th Sep 2024
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Hint: To prove the above two things we need to remember two basic things. First, triangles with the same base and between the same parallels are equal in area so area of Δ (ADC) = area of Δ (BDC) and second, two triangles are said to be similar if any two angles of both the triangles are equal. Using these two concepts we can easily solve the question.

Given:
A trapezium ABCD having its sides AB & DC parallel to each other. And its two diagonal AC & BD intersecting each other at O.
To prove:
Area of Δ (AOD) = Area of Δ (COB)
Δ AOB and Δ COD are similar.

To prove:
Area of Δ (AOD) = Area of Δ (COB)
First, we need to know that triangles with the same base and between the same parallels are equal in area so area of Δ (ADC) = area of Δ (BDC).
Now, if we deduct the area of Δ (ODC) from both the Δ (ADC) and Δ (BDC)
Area of Δ (ADC) - Area of Δ (ODC) = Area of Δ (BDC) - Area of Δ (ODC)
i.e. Area of Δ (AOD) = Area of Δ (COB)
Hence proved, Area of Δ (AOD) = Area of Δ (COB)
To prove:
Δ AOB and Δ COD are similar
We need to use the Angle-Angle similarity method according to which two triangles are said to be similar if any two angles of both the triangles are equal.
Now since $\angle AOB$ and $\angle DOC$ are vertically opposite angles,
$\Rightarrow \angle AOB = \angle DOC$
And, since AB ll DC with AC as traversal, $\angle OCD$ and $\angle OAB$ are alternate angles.
Hence,
$\Rightarrow \angle OCD = \angle OAB$
Hence proved, Δ AOB and Δ COD are similar triangles.