{"id":11380,"date":"2022-07-08T01:13:31","date_gmt":"2022-07-07T19:43:31","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11380"},"modified":"2022-07-08T01:15:09","modified_gmt":"2022-07-07T19:45:09","slug":"d-is-a-point-on-the-side-bc-of-triangle-abc-such-that-angle-adc-and-angle-bac-are-equal-prove-that-ca2-dc-times-cb","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/d-is-a-point-on-the-side-bc-of-triangle-abc-such-that-angle-adc-and-angle-bac-are-equal-prove-that-ca2-dc-times-cb\/","title":{"rendered":"D is a point on the side BC of \\(\\triangle\\) ABC such that \\(\\angle\\) ADC and \\(\\angle\\) BAC are equal. Prove that \\({CA}^2\\) = \\(DC \\times CB\\)."},"content":{"rendered":"

Solution :<\/h2>\n

Given<\/strong> : D is a point on the side BC of a triangle ABC, such that \\(\\angle\\) ADC = \\(\\angle\\) BAC\"triangles\"<\/p>\n

To Prove<\/strong> : \\({CA}^2\\) = \\(DC \\times CB\\)<\/p>\n

Proof<\/strong> : In triangles ABC and DAC,<\/p>\n

\\(\\angle\\) BAC = \\(\\angle\\) ADC\u00a0 \u00a0 \u00a0 \u00a0(given)<\/p>\n

\\(\\angle\\) C = \\(\\angle\\) C\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (common)<\/p>\n

\\(\\angle\\) ABC = \\(\\angle\\) DAC\u00a0 \u00a0 \u00a0 \u00a0 (third angles of the triangles)<\/p>\n

\\(\\therefore\\) \\(\\triangle\\)s ABC and DAC are equiangular and hence, similar<\/p>\n

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

\\(\\implies\\)\u00a0 \\({CA}^2\\) = \\(DC \\times CB\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Given : D is a point on the side BC of a triangle ABC, such that \\(\\angle\\) ADC = \\(\\angle\\) BAC To Prove : \\({CA}^2\\) = \\(DC \\times CB\\) Proof : In triangles ABC and DAC, \\(\\angle\\) BAC = \\(\\angle\\) ADC\u00a0 \u00a0 \u00a0 \u00a0(given) \\(\\angle\\) C = \\(\\angle\\) C\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …<\/p>\n

D is a point on the side BC of \\(\\triangle\\) ABC such that \\(\\angle\\) ADC and \\(\\angle\\) BAC are equal. Prove that \\({CA}^2\\) = \\(DC \\times CB\\).<\/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":"\nD is a point on the side BC of \\(\\triangle\\) ABC such that \\(\\angle\\) ADC and \\(\\angle\\) BAC are equal. Prove that \\({CA}^2\\) = \\(DC \\times CB\\). - 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\/d-is-a-point-on-the-side-bc-of-triangle-abc-such-that-angle-adc-and-angle-bac-are-equal-prove-that-ca2-dc-times-cb\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"D is a point on the side BC of \\(\\triangle\\) ABC such that \\(\\angle\\) ADC and \\(\\angle\\) BAC are equal. Prove that \\({CA}^2\\) = \\(DC \\times CB\\). - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : Given : D is a point on the side BC of a triangle ABC, such that (angle) ADC = (angle) BAC To Prove : ({CA}^2) = (DC times CB) Proof : In triangles ABC and DAC, (angle) BAC = (angle) ADC\u00a0 \u00a0 \u00a0 \u00a0(given) (angle) C = (angle) C\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 … D is a point on the side BC of (triangle) ABC such that (angle) ADC and (angle) BAC are equal. Prove that ({CA}^2) = (DC times CB). 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