{"id":11304,"date":"2022-07-01T03:31:09","date_gmt":"2022-06-30T22:01:09","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11304"},"modified":"2022-07-01T03:31:12","modified_gmt":"2022-06-30T22:01:12","slug":"the-diagonals-of-a-quadrilateral-abcd-intersect-each-other-at-the-point-o-such-that-aoover-bo-coover-do-show-that-abcd-is-a-trapezium","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/the-diagonals-of-a-quadrilateral-abcd-intersect-each-other-at-the-point-o-such-that-aoover-bo-coover-do-show-that-abcd-is-a-trapezium\/","title":{"rendered":"The diagonals of a quadrilateral ABCD intersect each other at the point O such that \\(AO\\over BO\\) = \\(CO\\over DO\\). Show that ABCD is a trapezium."},"content":{"rendered":"

Solution :<\/h2>\n

Given<\/strong> : A quadrilateral ABCD, in which the diagonals AC and BD intersect each other at O such that \"trapezium\" \\(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

The diagonals of a quadrilateral ABCD intersect each other at the point O such that \\(AO\\over BO\\) = \\(CO\\over DO\\). Show that ABCD is a trapezium.<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,912],"tags":[],"yoast_head":"\nThe diagonals of a quadrilateral ABCD intersect each other at the point O such that \\(AO\\over BO\\) = \\(CO\\over DO\\). 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