{"id":11479,"date":"2022-07-12T12:42:46","date_gmt":"2022-07-12T07:12:46","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11479"},"modified":"2022-07-12T12:53:02","modified_gmt":"2022-07-12T07:23:02","slug":"in-the-figure-o-is-a-point-in-the-interior-of-a-triangle-abc-od-perp-bc-oe-perp-ac-and-of-perp-ab","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-the-figure-o-is-a-point-in-the-interior-of-a-triangle-abc-od-perp-bc-oe-perp-ac-and-of-perp-ab\/","title":{"rendered":"In the figure, O is a point in the interior of a triangle ABC, OD \\(\\perp\\) BC, OE \\(\\perp\\) AC and OF \\(\\perp\\) AB."},"content":{"rendered":"

Question :<\/h2>\n

In the figure, O is a point in the interior of a triangle ABC, OD \\(\\perp\\) BC, OE \\(\\perp\\) AC and OF \\(\\perp\\) AB. Show that\"triangles\"<\/p>\n

(i)<\/strong>\u00a0 \\({OA}^2\\) + \\({OB}^2\\) + \\({OC}^2\\) – \\({OD}^2\\) – \\({OE}^2\\) – \\({OF}^2\\) = \\({AF}^2\\) + \\({BD}^2\\) + \\({CE}^2\\)<\/p>\n

(ii)<\/strong>\u00a0 \\({AF}^2\\) + \\({BD}^2\\) + \\({CE}^2\\) = \\({AE}^2\\) + \\({CD}^2\\) + \\({BF}^2\\)<\/p>\n

Solution :<\/h2>\n

Join AO, BO and CO as shown in fig.<\/p>\n

(i)<\/strong>\u00a0 In right \\(\\triangle\\)s OFA, ODB and OEC, we have<\/p>\n

\\({OA}^2\\) = \\({AF}^2\\) + \\({OF}^2\\)<\/p>\n

\\({OB}^2\\) = \\({BD}^2\\) + \\({OD}^2\\)<\/p>\n

and \\({OC}^2\\) = \\({CE}^2\\) + \\({OE}^2\\)<\/p>\n

Adding all these, we get<\/p>\n

\\({OA}^2\\) + \\({OB}^2\\) + \\({OC}^2\\) = \\({AF}^2\\) + \\({BD}^2\\) + \\({CE}^2\\) + \\({OD}^2\\) + \\({OE}^2\\) + \\({OF}^2\\)<\/p>\n

or\u00a0 \u00a0\\({OA}^2\\) + \\({OB}^2\\) + \\({OC}^2\\) – \\({OD}^2\\) – \\({OE}^2\\) – \\({OF}^2\\) = \\({AF}^2\\) + \\({BD}^2\\) + \\({CE}^2\\)<\/strong><\/p>\n

(ii)\u00a0 In right \\(\\triangle\\)s ODB and ODC, we have :<\/p>\n

\\({OB}^2\\) = \\({BD}^2\\) + \\({OD}^2\\)\u00a0 and\u00a0 \\({OC}^2\\) = \\({OD}^2\\) + \\({CD}^2\\)<\/p>\n

or\u00a0 \\({OB}^2\\) – \\({OC}^2\\) = \\({BD}^2\\) – \\({CD}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……….(1)<\/p>\n

Similarly, we have :<\/p>\n

\\({OC}^2\\) – \\({OA}^2\\) = \\({CE}^2\\) – \\({AE}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………..(2)<\/p>\n

\\({OA}^2\\) – \\({OB}^2\\) = \\({AF}^2\\) + \\({BF}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(3)<\/p>\n

Adding equations (1), (2) and (3), we get<\/p>\n

\\({AF}^2\\) + \\({BD}^2\\) + \\({CE}^2\\) = \\({AE}^2\\) + \\({CD}^2\\) + \\({BF}^2\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Question : In the figure, O is a point in the interior of a triangle ABC, OD \\(\\perp\\) BC, OE \\(\\perp\\) AC and OF \\(\\perp\\) AB. Show that (i)\u00a0 \\({OA}^2\\) + \\({OB}^2\\) + \\({OC}^2\\) – \\({OD}^2\\) – \\({OE}^2\\) – \\({OF}^2\\) = \\({AF}^2\\) + \\({BD}^2\\) + \\({CE}^2\\) (ii)\u00a0 \\({AF}^2\\) + \\({BD}^2\\) + \\({CE}^2\\) = \\({AE}^2\\) + …<\/p>\n

In the figure, O is a point in the interior of a triangle ABC, OD \\(\\perp\\) BC, OE \\(\\perp\\) AC and OF \\(\\perp\\) AB.<\/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, O is a point in the interior of a triangle ABC, OD \\(\\perp\\) BC, OE \\(\\perp\\) AC and OF \\(\\perp\\) AB. - 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-o-is-a-point-in-the-interior-of-a-triangle-abc-od-perp-bc-oe-perp-ac-and-of-perp-ab\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"In the figure, O is a point in the interior of a triangle ABC, OD \\(\\perp\\) BC, OE \\(\\perp\\) AC and OF \\(\\perp\\) AB. - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Question : In the figure, O is a point in the interior of a triangle ABC, OD (perp) BC, OE (perp) AC and OF (perp) AB. Show that (i)\u00a0 ({OA}^2) + ({OB}^2) + ({OC}^2) – ({OD}^2) – ({OE}^2) – ({OF}^2) = ({AF}^2) + ({BD}^2) + ({CE}^2) (ii)\u00a0 ({AF}^2) + ({BD}^2) + ({CE}^2) = ({AE}^2) + … In the figure, O is a point in the interior of a triangle ABC, OD (perp) BC, OE (perp) AC and OF (perp) AB. 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