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प्रश्न
Prove that the diagonals of a trapezium divide each other proportionally.
बेरीज
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उत्तर
Given: In trapezium ABCD, AB || CD. Diagonals AC and BD intersect at O.
In △AOB and △COD:
- ∠ABO = ∠CDO (alternate interior angles, since AB || CD)
- ∠AOB = ∠COD (vertically opposite angles)
⇒ △AOB ∼ △COD (by AA similarity)
Therefore, corresponding sides are proportional:
`(AO)/(OC) = (BO)/(OD)`
Hence proved that the diagonals of a trapezium divide each other proportionally.
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