{"id":11334,"date":"2022-07-07T19:46:11","date_gmt":"2022-07-07T14:16:11","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11334"},"modified":"2022-07-07T19:46:15","modified_gmt":"2022-07-07T14:16:15","slug":"diagonals-ac-and-bd-of-a-trapezium-abcd-with-ab-dc-intersect-each-other-at-the-point-o-using-a-similarity-criterion-for-two-triangles-show-that-oaover-oc-obover-od","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/diagonals-ac-and-bd-of-a-trapezium-abcd-with-ab-dc-intersect-each-other-at-the-point-o-using-a-similarity-criterion-for-two-triangles-show-that-oaover-oc-obover-od\/","title":{"rendered":"Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \\(OA\\over OC\\) = \\(OB\\over OD\\)."},"content":{"rendered":"

Solution :<\/h2>\n

Given<\/strong> : A trapezium ABC whose diagonals AC and BD intersect each other at O and AB || DC.\"trapezium\"<\/p>\n

To Prove<\/strong> : \\(OA\\over OC\\) = \\(OB\\over OD\\)<\/p>\n

Proof<\/strong> : In triangle DOC and AOB,<\/p>\n

AB || DC and AC is transversal, then<\/p>\n

\\(\\angle\\) DCO = \\(\\angle\\) OAB\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(Alternate angles)<\/p>\n

\\(\\angle\\) ODC = \\(\\angle\\) OBA\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (Alternate angles)<\/p>\n

\\(\\angle\\) DOC = \\(\\angle\\) AOB\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(vertically opposite angles)<\/p>\n

Hence, By AAA similarity,<\/p>\n

\\(\\triangle\\) DOC ~ \\(\\triangle\\) BOA<\/p>\n

\\(\\implies\\)\u00a0 \u00a0 \\(OA\\over OC\\) = \\(OB\\over OD\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Given : A trapezium ABC whose diagonals AC and BD intersect each other at O and AB || DC. To Prove : \\(OA\\over OC\\) = \\(OB\\over OD\\) Proof : In triangle DOC and AOB, AB || DC and AC is transversal, then \\(\\angle\\) DCO = \\(\\angle\\) OAB\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …<\/p>\n

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \\(OA\\over OC\\) = \\(OB\\over OD\\).<\/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":"\nDiagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \\(OA\\over OC\\) = \\(OB\\over OD\\). - 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\/diagonals-ac-and-bd-of-a-trapezium-abcd-with-ab-dc-intersect-each-other-at-the-point-o-using-a-similarity-criterion-for-two-triangles-show-that-oaover-oc-obover-od\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \\(OA\\over OC\\) = \\(OB\\over OD\\). - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : Given : A trapezium ABC whose diagonals AC and BD intersect each other at O and AB || DC. 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