![\"trapezium\"](\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-2-10.jpeg?resize=245%2C114&ssl=1\")
\\(AO\\over BO\\) = \\(CO\\over DO\\).<\/p>\nGiven<\/strong> : A quadrilateral ABCD, in which the diagonals AC and BD intersect each other at O such that \\(AO\\over BO\\) = \\(CO\\over DO\\).<\/p>\n
To\u00a0 Prove<\/strong> : ABCD is a trapezium.<\/p>\n
Construction<\/strong> : Draw EO || BA, meeting AD in E.<\/p>\n
Proof<\/strong> : In triangle ABD, EO || BA<\/p>\n
By basic proportionality theorem,<\/p>\n
\\(DE\\over EA\\) = \\(DO\\over OB\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …………(1)<\/p>\n
But Given that, \\(AO\\over BO\\) = \\(CO\\over DO\\) \\(\\implies\\)\u00a0 \\(DO\\over BO\\) = \\(CO\\over AO\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0…………..(2)<\/p>\n
From (1) and (2), we get<\/p>\n
\\(DE\\over EA\\) = \\(CO\\over OA\\)<\/p>\n
By converse of basic proportionality theorem,<\/p>\n
\\(\\implies\\) EO || DC<\/p>\n
But by construction, EO || BA<\/p>\n
\\(\\therefore\\)\u00a0 DC || BA<\/p>\n
Hence, ABCD is a trapezium.<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"
Solution : Given : A quadrilateral ABCD, in which the diagonals AC and BD intersect each other at O such that \\(AO\\over BO\\) = \\(CO\\over DO\\). To\u00a0 Prove : ABCD is a trapezium. Construction : Draw EO || BA, meeting AD in E. Proof : In triangle ABD, EO || BA By basic proportionality theorem, …<\/p>\n