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In the figure, diagonals AC and BD of a trapezium ABCD with AB//DC intersect each other at ‘O’. Prove that ar (ΔAOD) = ar (ΔBOC) |
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Answer» Given that AB // CD Now ΔADC and ΔBCD are on the same base and between the same parallels AB // CD. ∴ ΔADC = ΔBCD ⇒ ΔADC – ΔCOD = ΔBCD – ΔCOD ⇒ ΔAOD = ΔBOC [from the figure] |
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