{"id":11544,"date":"2022-07-18T14:40:47","date_gmt":"2022-07-18T09:10:47","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11544"},"modified":"2022-07-18T14:40:52","modified_gmt":"2022-07-18T09:10:52","slug":"in-the-figure-abc-is-a-triangle-in-which-angle-abc-90-and-ad-perp-cb-produced-prove-that-ac2-ab2-bc2-2bc-bd","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-the-figure-abc-is-a-triangle-in-which-angle-abc-90-and-ad-perp-cb-produced-prove-that-ac2-ab2-bc2-2bc-bd\/","title":{"rendered":"In the figure, ABC is a triangle in which \\(\\angle\\) ABC > 90 and AD \\(\\perp\\) CB produced. Prove that \\({AC}^2\\) = \\({AB}^2\\) + \\({BC}^2\\) + 2BC.BD."},"content":{"rendered":"

Solution :<\/h2>\n

Given<\/strong> : ABC is triangle in which \\(\\angle\\) ABC > 90 and AD \\(\\perp\\) CB produced.\"triangle\"<\/p>\n

To Prove<\/strong> : \\({AC}^2\\) = \\({AB}^2\\) + \\({BC}^2\\) + 2BC.BD<\/p>\n

Proof<\/strong> : Since \\(\\triangle\\) ADB is a right triangle, angled at D. Therefore, by Pythagoras theorem,<\/p>\n

\\({AB}^2\\) = \\({AD}^2\\) + \\({DB}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……..(1)<\/p>\n

Again, in triangle ADC is a right triangle, angled at D.<\/p>\n

Therefore, by Pythagoras theorem, we have :<\/p>\n

\\({AC}^2\\) = \\({AD}^2\\) + \\({DC}^2\\)<\/p>\n

\\(\\implies\\)\u00a0 \\({AC}^2\\) = \\({AD}^2\\) + \\({DB + BC}^2\\)<\/p>\n

\\(\\implies\\)\u00a0 \\({AC}^2\\) = \\({AD}^2\\) + \\({DB}^2\\) + \\({BC}^2\\) + 2BC.BD<\/p>\n

\\(\\implies\\)\u00a0 \\({AC}^2\\) = \\({AB}^2\\) + \\({BC}^2\\) + 2BC.BD\u00a0<\/strong> \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(by using 1)<\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Given : ABC is triangle in which \\(\\angle\\) ABC > 90 and AD \\(\\perp\\) CB produced. To Prove : \\({AC}^2\\) = \\({AB}^2\\) + \\({BC}^2\\) + 2BC.BD Proof : Since \\(\\triangle\\) ADB is a right triangle, angled at D. Therefore, by Pythagoras theorem, \\({AB}^2\\) = \\({AD}^2\\) + \\({DB}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……..(1) Again, …<\/p>\n

In the figure, ABC is a triangle in which \\(\\angle\\) ABC > 90 and AD \\(\\perp\\) CB produced. Prove that \\({AC}^2\\) = \\({AB}^2\\) + \\({BC}^2\\) + 2BC.BD.<\/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 is a triangle in which \\(\\angle\\) ABC > 90 and AD \\(\\perp\\) CB produced. Prove that \\({AC}^2\\) = \\({AB}^2\\) + \\({BC}^2\\) + 2BC.BD. - 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-is-a-triangle-in-which-angle-abc-90-and-ad-perp-cb-produced-prove-that-ac2-ab2-bc2-2bc-bd\/\" \/>\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 is a triangle in which \\(\\angle\\) ABC > 90 and AD \\(\\perp\\) CB produced. Prove that \\({AC}^2\\) = \\({AB}^2\\) + \\({BC}^2\\) + 2BC.BD. - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : Given : ABC is triangle in which (angle) ABC > 90 and AD (perp) CB produced. To Prove : ({AC}^2) = ({AB}^2) + ({BC}^2) + 2BC.BD Proof : Since (triangle) ADB is a right triangle, angled at D. Therefore, by Pythagoras theorem, ({AB}^2) = ({AD}^2) + ({DB}^2)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……..(1) Again, … In the figure, ABC is a triangle in which (angle) ABC > 90 and AD (perp) CB produced. Prove that ({AC}^2) = ({AB}^2) + ({BC}^2) + 2BC.BD. 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