{"id":11452,"date":"2022-07-11T16:42:51","date_gmt":"2022-07-11T11:12:51","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11452"},"modified":"2022-07-11T16:42:56","modified_gmt":"2022-07-11T11:12:56","slug":"in-the-figure-abd-is-triangle-right-angled-at-a-and-ac-perp-bd-show-that-i-ab2-bc-bd-ii-ac2-bc-dc-iii-ad2-bd-cd","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-the-figure-abd-is-triangle-right-angled-at-a-and-ac-perp-bd-show-that-i-ab2-bc-bd-ii-ac2-bc-dc-iii-ad2-bd-cd\/","title":{"rendered":"In the figure, ABD is triangle right angled at A and AC \\(\\perp\\) BD. Show that (i) \\({AB}^2\\) = BC.BD (ii) \\({AC}^2\\) = BC.DC (iii) \\({AD}^2\\) = BD.CD"},"content":{"rendered":"

Solution :<\/h2>\n

(i)<\/strong> Since, AC \\(\\perp\\) BD, therefore,\"triangle\"<\/p>\n

\\(\\triangle\\) ABC ~ \\(\\triangle\\) DBA\u00a0 and\u00a0 each triangle is similar to whole triangle ABD.<\/p>\n

\\(\\implies\\)\u00a0 \\(AB\\over BD\\) = \\(BC\\over AB\\)<\/p>\n

So, \\({AB}^2\\) = BC.BD<\/strong><\/p>\n

(ii)<\/strong> Since, \\(\\triangle\\) ABC ~ \\(\\triangle\\) DAC, therefore,<\/p>\n

\\(\\implies\\)\u00a0 \\(AC\\over BC\\) = \\(DC\\over AC\\)<\/p>\n

So,\u00a0\\({AC}^2\\) = BC.DC<\/strong><\/p>\n

(iii)<\/strong> Since, \\(\\triangle\\) ACD ~ \\(\\triangle\\) BAD, therefore,<\/p>\n

\\(\\implies\\)\u00a0 \\(AD\\over CD\\) = \\(BD\\over AD\\)<\/p>\n

So,\u00a0\\({AD}^2\\) = BD.CD<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : (i) Since, AC \\(\\perp\\) BD, therefore, \\(\\triangle\\) ABC ~ \\(\\triangle\\) DBA\u00a0 and\u00a0 each triangle is similar to whole triangle ABD. \\(\\implies\\)\u00a0 \\(AB\\over BD\\) = \\(BC\\over AB\\) So, \\({AB}^2\\) = BC.BD (ii) Since, \\(\\triangle\\) ABC ~ \\(\\triangle\\) DAC, therefore, \\(\\implies\\)\u00a0 \\(AC\\over BC\\) = \\(DC\\over AC\\) So,\u00a0\\({AC}^2\\) = BC.DC (iii) Since, \\(\\triangle\\) ACD ~ \\(\\triangle\\) …<\/p>\n

In the figure, ABD is triangle right angled at A and AC \\(\\perp\\) BD. Show that (i) \\({AB}^2\\) = BC.BD (ii) \\({AC}^2\\) = BC.DC (iii) \\({AD}^2\\) = BD.CD<\/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, ABD is triangle right angled at A and AC \\(\\perp\\) BD. Show that (i) \\({AB}^2\\) = BC.BD (ii) \\({AC}^2\\) = BC.DC (iii) \\({AD}^2\\) = BD.CD - 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-abd-is-triangle-right-angled-at-a-and-ac-perp-bd-show-that-i-ab2-bc-bd-ii-ac2-bc-dc-iii-ad2-bd-cd\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"In the figure, ABD is triangle right angled at A and AC \\(\\perp\\) BD. Show that (i) \\({AB}^2\\) = BC.BD (ii) \\({AC}^2\\) = BC.DC (iii) \\({AD}^2\\) = BD.CD - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : (i) Since, AC (perp) BD, therefore, (triangle) ABC ~ (triangle) DBA\u00a0 and\u00a0 each triangle is similar to whole triangle ABD. (implies)\u00a0 (ABover BD) = (BCover AB) So, ({AB}^2) = BC.BD (ii) Since, (triangle) ABC ~ (triangle) DAC, therefore, (implies)\u00a0 (ACover BC) = (DCover AC) So,\u00a0({AC}^2) = BC.DC (iii) Since, (triangle) ACD ~ (triangle) … In the figure, ABD is triangle right angled at A and AC (perp) BD. 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