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Question
In the given figure, the diagonals AC and BD intersect at point O. If OB = OD and AB//DC,
show that:
(i) Area (Δ DOC) = Area (Δ AOB).
(ii) Area (Δ DCB) = Area (Δ ACB).
(iii) ABCD is a parallelogram.
Solution
(i) The ratio of the area of triangles with the same vertex and bases along the same line is equal to the ratio of their respective bases. So, we have:
`["Area of "Δ"DOC"]/["Area of" Δ"BOC" ] = [DO]/[BO]` = 1 .....(i)
Similarly
`["Area of "Δ"DOA"]/["Area of" Δ"BOA" ] = [DO]/[BO]` = 1 ......(ii)
We know that the area of triangles on the same base and between the same parallel lines are equal.
Area of Δ ACD = Area of Δ BCD
Area of Δ AOD + Area of Δ DOC = Area of Δ DOC + Area of Δ BOC
⇒ Area of Δ AOD = Area of Δ BOC .....(iii)
From 1, 2 and 3 we have
Area (Δ DOC) = Area (Δ AOB)
Hence Proved.
(ii) Similarly, from 1, 2 and 3, we also have
Area of Δ DCB = Area of Δ DOC + Area of Δ BOC = Area of Δ AOB + Area of Δ BOC = Area of Δ ABC
So Area of Δ DCB = Area of Δ ABC
Hence Proved.
(iii) We know that the area of triangles on the same base and between the same parallel lines are equal.
Given: triangles are equal in the area on the common base, so it indicates AD || BC.
So, ABCD is a parallelogram.
Hence Proved.
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