{"id":11301,"date":"2022-07-01T03:21:00","date_gmt":"2022-06-30T21:51:00","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11301"},"modified":"2022-07-01T03:21:03","modified_gmt":"2022-06-30T21:51:03","slug":"abcd-is-a-trapezium-in-which-ab-dc-and-its-diagonals-intersect-each-other-at-the-point-o-show-that-aoover-bo-coover-do","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/abcd-is-a-trapezium-in-which-ab-dc-and-its-diagonals-intersect-each-other-at-the-point-o-show-that-aoover-bo-coover-do\/","title":{"rendered":"ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \\(AO\\over BO\\) = \\(CO\\over DO\\)."},"content":{"rendered":"

Solution :<\/h2>\n

Given<\/strong> : A trapezium ABCD, in which the diagonals AC and BD intersect each other at O.\"trapezium\"<\/p>\n

To\u00a0 Prove<\/strong> : \\(AO\\over BO\\) = \\(CO\\over DO\\)<\/p>\n

Construction<\/strong> : Draw EF || BA || CD, meeting AD in E.<\/p>\n

Proof<\/strong> : In triangle ABD, EF || AB<\/p>\n

By basic proportionality theorem,<\/p>\n

\\(DO\\over OB\\) = \\(DE\\over AE\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …………(1)<\/p>\n

In triangle CDA, EO || DC,<\/p>\n

By basic proportionality theorem,<\/p>\n

\\(CO\\over OA\\) = \\(DE\\over AE\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …………..(2)<\/p>\n

From (1) and (2), we get<\/p>\n

\\(DO\\over OB\\) = \\(CO\\over OA\\)<\/p>\n

\\(\\implies\\)\u00a0 \u00a0 \\(AO\\over BO\\) = \\(CO\\over DO\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Given : A trapezium ABCD, in which the diagonals AC and BD intersect each other at O. To\u00a0 Prove : \\(AO\\over BO\\) = \\(CO\\over DO\\) Construction : Draw EF || BA || CD, meeting AD in E. Proof : In triangle ABD, EF || AB By basic proportionality theorem, \\(DO\\over OB\\) = \\(DE\\over …<\/p>\n

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \\(AO\\over BO\\) = \\(CO\\over DO\\).<\/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":"\nABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \\(AO\\over BO\\) = \\(CO\\over DO\\). - Mathemerize<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/abcd-is-a-trapezium-in-which-ab-dc-and-its-diagonals-intersect-each-other-at-the-point-o-show-that-aoover-bo-coover-do\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \\(AO\\over BO\\) = \\(CO\\over DO\\). - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : Given : A trapezium ABCD, in which the diagonals AC and BD intersect each other at O. To\u00a0 Prove : (AOover BO) = (COover DO) Construction : Draw EF || BA || CD, meeting AD in E. Proof : In triangle ABD, EF || AB By basic proportionality theorem, (DOover OB) = (DEover … ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that (AOover BO) = (COover DO). 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