{"id":11407,"date":"2022-07-08T22:27:12","date_gmt":"2022-07-08T16:57:12","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11407"},"modified":"2022-07-08T22:27:15","modified_gmt":"2022-07-08T16:57:15","slug":"in-the-figure-abc-and-dbc-are-two-triangles-on-the-same-base-bc-if-ad-intersects-bc-at-o-show-that-areatriangle-abcover-areatriangle-dbc-aoover-do","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-the-figure-abc-and-dbc-are-two-triangles-on-the-same-base-bc-if-ad-intersects-bc-at-o-show-that-areatriangle-abcover-areatriangle-dbc-aoover-do\/","title":{"rendered":"In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \\(area(\\triangle ABC)\\over area(\\triangle DBC)\\) = \\(AO\\over DO\\)."},"content":{"rendered":"

Solution :<\/h2>\n

Given<\/strong> : Two triangles ABC and DBC which stand on the same base BC but on the opposite sides of BC.\"trapezium\"<\/p>\n

To Prove<\/strong> : \\(area(\\triangle ABC)\\over area(\\triangle DBC)\\) = \\(AO\\over DO\\)<\/p>\n

Construction<\/strong> : Draw AE \\(\\perp\\) BC and DF \\(\\perp\\) BC.<\/p>\n

Proof<\/strong> : In triangles AOE and DOF, we have :<\/p>\n

\\(\\angle\\) AEO = \\(\\angle\\) DFO = 90<\/p>\n

\\(\\angle\\) AOE = \\(\\angle\\) DOF\u00a0 \u00a0 \u00a0 \u00a0(vertically opposite angles)<\/p>\n

By AA similarity, we have :<\/p>\n

\\(\\triangle\\) AOE ~ \\(\\triangle\\) DOF<\/p>\n

So, \\(AE\\over DF\\) = \\(AO\\over OD\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……..(1)<\/p>\n

Now, \\(area(\\triangle ABC)\\over area(\\triangle DBC)\\) = \\({1\\over 2} \\times BC \\times AE\\over {1\\over 2}\\times BC \\times DF\\) = \\(AE\\over DF\\) = \\(AO\\over OD\\)<\/p>\n

\\(\\therefore\\)\u00a0 \u00a0 \u00a0\\(area(\\triangle ABC)\\over area(\\triangle DBC)\\) = \\(AO\\over OD\\)<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Given : Two triangles ABC and DBC which stand on the same base BC but on the opposite sides of BC. To Prove : \\(area(\\triangle ABC)\\over area(\\triangle DBC)\\) = \\(AO\\over DO\\) Construction : Draw AE \\(\\perp\\) BC and DF \\(\\perp\\) BC. Proof : In triangles AOE and DOF, we have : \\(\\angle\\) AEO …<\/p>\n

In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \\(area(\\triangle ABC)\\over area(\\triangle DBC)\\) = \\(AO\\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":"\nIn the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \\(area(\\triangle ABC)\\over area(\\triangle DBC)\\) = \\(AO\\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\/in-the-figure-abc-and-dbc-are-two-triangles-on-the-same-base-bc-if-ad-intersects-bc-at-o-show-that-areatriangle-abcover-areatriangle-dbc-aoover-do\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \\(area(\\triangle ABC)\\over area(\\triangle DBC)\\) = \\(AO\\over DO\\). - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : Given : Two triangles ABC and DBC which stand on the same base BC but on the opposite sides of BC. 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