{"id":11364,"date":"2022-07-07T23:22:03","date_gmt":"2022-07-07T17:52:03","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11364"},"modified":"2022-07-07T23:22:53","modified_gmt":"2022-07-07T17:52:53","slug":"in-the-figure-abc-and-amp-are-two-right-triangles-right-angled-at-b-and-m-respectively-prove-that-i-triangle-abc-triangle-amp-ii-caover-pa-bcover-mp","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-the-figure-abc-and-amp-are-two-right-triangles-right-angled-at-b-and-m-respectively-prove-that-i-triangle-abc-triangle-amp-ii-caover-pa-bcover-mp\/","title":{"rendered":"In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, Prove that (i) \\(\\triangle\\) ABC ~ \\(\\triangle\\)  AMP    (ii) \\(CA\\over PA\\) = \\(BC\\over MP\\)"},"content":{"rendered":"<h2>Solution :<\/h2>\n<p><strong>(i)<\/strong>\u00a0 In triangles, ABC and AMP, we have<img decoding=\"async\" loading=\"lazy\" class=\" wp-image-11366 alignright\" src=\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-3-9.jpeg?resize=234%2C171&#038;ssl=1\" alt=\"triangles\" width=\"234\" height=\"171\" srcset=\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-3-9.jpeg?resize=300%2C219&amp;ssl=1 300w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-3-9.jpeg?resize=1024%2C747&amp;ssl=1 1024w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-3-9.jpeg?resize=768%2C560&amp;ssl=1 768w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-3-9.jpeg?resize=1536%2C1120&amp;ssl=1 1536w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-3-9.jpeg?w=1700&amp;ssl=1 1700w\" sizes=\"(max-width: 234px) 100vw, 234px\" data-recalc-dims=\"1\" \/><\/p>\n<p>\\(\\angle\\) ABC = \\(\\angle\\) AMP = 90\u00a0 \u00a0 \u00a0 \u00a0(given)<\/p>\n<p>\\(\\angle\\)\u00a0 BAC = \\(\\angle\\) MAP\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (common angles)<\/p>\n<p>\\(\\therefore\\)\u00a0 By AA similarity, we have<\/p>\n<p><strong>\\(\\triangle\\) ABC ~ \\(\\triangle\\) AMP<\/strong><\/p>\n<p><strong>(ii)<\/strong>\u00a0 We have :<\/p>\n<p>\\(\\triangle\\) ABC ~ \\(\\triangle\\) AMP\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (as proved above)<\/p>\n<p>So, In similar triangles, corresponding sides are proportional.<\/p>\n<p>\\(\\implies\\)<strong>\u00a0 \u00a0\\(CA\\over PA\\) = \\(BC\\over MP\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Solution : (i)\u00a0 In triangles, ABC and AMP, we have \\(\\angle\\) ABC = \\(\\angle\\) AMP = 90\u00a0 \u00a0 \u00a0 \u00a0(given) \\(\\angle\\)\u00a0 BAC = \\(\\angle\\) MAP\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (common angles) \\(\\therefore\\)\u00a0 By AA similarity, we have \\(\\triangle\\) ABC ~ \\(\\triangle\\) AMP (ii)\u00a0 We have : \\(\\triangle\\) ABC ~ \\(\\triangle\\) AMP\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathemerize.com\/in-the-figure-abc-and-amp-are-two-right-triangles-right-angled-at-b-and-m-respectively-prove-that-i-triangle-abc-triangle-amp-ii-caover-pa-bcover-mp\/\"> <span class=\"screen-reader-text\">In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, Prove that (i) \\(\\triangle\\) ABC ~ \\(\\triangle\\)  AMP    (ii) \\(CA\\over PA\\) = \\(BC\\over MP\\)<\/span> Read More &raquo;<\/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":"<!-- This site is optimized with the Yoast SEO plugin v20.12 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, Prove that (i) \\(\\triangle\\) ABC ~ \\(\\triangle\\) AMP  (ii) \\(CA\\over PA\\) = \\(BC\\over MP\\) - 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-amp-are-two-right-triangles-right-angled-at-b-and-m-respectively-prove-that-i-triangle-abc-triangle-amp-ii-caover-pa-bcover-mp\/\" \/>\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 AMP are two right triangles, right angled at B and M respectively, Prove that (i) \\(\\triangle\\) ABC ~ \\(\\triangle\\) AMP  (ii) \\(CA\\over PA\\) = \\(BC\\over MP\\) - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : (i)\u00a0 In triangles, ABC and AMP, we have (angle) ABC = (angle) AMP = 90\u00a0 \u00a0 \u00a0 \u00a0(given) (angle)\u00a0 BAC = (angle) MAP\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (common angles) (therefore)\u00a0 By AA similarity, we have (triangle) ABC ~ (triangle) AMP (ii)\u00a0 We have : (triangle) ABC ~ (triangle) AMP\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &hellip; In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, Prove that (i) (triangle) ABC ~ (triangle) AMP  (ii) (CAover PA) = (BCover MP) Read More &raquo;\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mathemerize.com\/in-the-figure-abc-and-amp-are-two-right-triangles-right-angled-at-b-and-m-respectively-prove-that-i-triangle-abc-triangle-amp-ii-caover-pa-bcover-mp\/\" \/>\n<meta property=\"og:site_name\" content=\"Mathemerize\" \/>\n<meta property=\"article:published_time\" content=\"2022-07-07T17:52:03+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2022-07-07T17:52:53+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-3-9-300x219.jpeg\" \/>\n<meta name=\"author\" content=\"mathemerize\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"mathemerize\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/mathemerize.com\/in-the-figure-abc-and-amp-are-two-right-triangles-right-angled-at-b-and-m-respectively-prove-that-i-triangle-abc-triangle-amp-ii-caover-pa-bcover-mp\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/mathemerize.com\/in-the-figure-abc-and-amp-are-two-right-triangles-right-angled-at-b-and-m-respectively-prove-that-i-triangle-abc-triangle-amp-ii-caover-pa-bcover-mp\/\"},\"author\":{\"name\":\"mathemerize\",\"@id\":\"https:\/\/mathemerize.com\/#\/schema\/person\/104c8bc54f90618130a6665299bc55df\"},\"headline\":\"In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, Prove that (i) \\\\(\\\\triangle\\\\) ABC ~ \\\\(\\\\triangle\\\\) AMP (ii) \\\\(CA\\\\over PA\\\\) = \\\\(BC\\\\over MP\\\\)\",\"datePublished\":\"2022-07-07T17:52:03+00:00\",\"dateModified\":\"2022-07-07T17:52:53+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/mathemerize.com\/in-the-figure-abc-and-amp-are-two-right-triangles-right-angled-at-b-and-m-respectively-prove-that-i-triangle-abc-triangle-amp-ii-caover-pa-bcover-mp\/\"},\"wordCount\":86,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/mathemerize.com\/#organization\"},\"articleSection\":[\"Maths Questions\",\"Similar Triangles Questions\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/mathemerize.com\/in-the-figure-abc-and-amp-are-two-right-triangles-right-angled-at-b-and-m-respectively-prove-that-i-triangle-abc-triangle-amp-ii-caover-pa-bcover-mp\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/mathemerize.com\/in-the-figure-abc-and-amp-are-two-right-triangles-right-angled-at-b-and-m-respectively-prove-that-i-triangle-abc-triangle-amp-ii-caover-pa-bcover-mp\/\",\"url\":\"https:\/\/mathemerize.com\/in-the-figure-abc-and-amp-are-two-right-triangles-right-angled-at-b-and-m-respectively-prove-that-i-triangle-abc-triangle-amp-ii-caover-pa-bcover-mp\/\",\"name\":\"In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, Prove that (i) \\\\(\\\\triangle\\\\) ABC ~ \\\\(\\\\triangle\\\\) AMP (ii) \\\\(CA\\\\over PA\\\\) = \\\\(BC\\\\over MP\\\\) - 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