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Question
In a trapezium ABCD, AB || DC and its diagonals intersect at O. Prove that `(OA)/(OC) = (OB)/(OD)`.
Theorem
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Solution

Given: In a trapezium ABCD, AB || DC.
Also, AC and BD intersect each other at O.
Construction: Draw XY || AB and CD through ‘Ο’.
To prove: `(OA)/(OC) = (OB)/(OD)`
Proof: In ΔABC, OY || AB
`(BY)/(YC) = "AO"/"OC"` ...(1) (By BPT)
In ΔBCD, OY || CD
`(BY)/(YC) = (OB)/(OD)` ...(2) (By BPT)
From equation (1) and (2)
`(OA)/(OC) = (OB)/(OD)`
Hence proved.
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