{"id":11556,"date":"2022-07-18T14:48:11","date_gmt":"2022-07-18T09:18:11","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11556"},"modified":"2022-07-18T14:48:14","modified_gmt":"2022-07-18T09:18:14","slug":"prove-that-the-sum-of-the-squares-of-the-diagonals-of-a-parallelogram-is-equal-to-the-sum-of-the-squares-of-its-sides","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/prove-that-the-sum-of-the-squares-of-the-diagonals-of-a-parallelogram-is-equal-to-the-sum-of-the-squares-of-its-sides\/","title":{"rendered":"Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides."},"content":{"rendered":"<h2>Solution :<\/h2>\n<p>We know that AD is a median of triangle ABC, then<img decoding=\"async\" loading=\"lazy\" class=\" wp-image-11579 alignright\" src=\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-6-6.jpeg?resize=162%2C127&#038;ssl=1\" alt=\"\" width=\"162\" height=\"127\" srcset=\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-6-6.jpeg?resize=300%2C235&amp;ssl=1 300w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-6-6.jpeg?resize=1024%2C804&amp;ssl=1 1024w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-6-6.jpeg?resize=768%2C603&amp;ssl=1 768w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-6-6.jpeg?resize=1536%2C1205&amp;ssl=1 1536w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-6-6.jpeg?w=1700&amp;ssl=1 1700w\" sizes=\"(max-width: 162px) 100vw, 162px\" data-recalc-dims=\"1\" \/><\/p>\n<p>\\({AB}^2\\) + \\({AC}^2\\) = 2\\({AD}^2\\) + \\({1\\over 2}{BC}^2\\)<\/p>\n<p>Since the diagonals of a parallelogram bisect each other, therefore BO and DO are medians of triangles ABC and ADC respectively.<\/p>\n<p>\\(\\therefore\\)\u00a0 \\({AB}^2\\) + \\({BC}^2\\) = 2\\({BO}^2\\) + \\({1\\over 2}{AC}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;&#8230;&#8230;..(1)<\/p>\n<p>and\u00a0 \u00a0 \u00a0 \\({AD}^2\\) + \\({CD}^2\\) = 2\\({DO}^2\\) + \\({1\\over 2}{AC}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;&#8230;&#8230;&#8230;(2)<\/p>\n<p>Adding (1) and (2), we get<\/p>\n<p>\\({AB}^2\\) + \\({BC}^2\\) + \\({AD}^2\\) + \\({CD}^2\\) = 2(\\({BO}^2\\) + \\({DO}^2\\)) + \\({AC}^2\\)<\/p>\n<p>\\(\\implies\\)\u00a0 \\({AB}^2\\) + \\({BC}^2\\) + \\({AD}^2\\) + \\({CD}^2\\) = 2(\\({1\\over 4}{BD}^2\\) + \\({1\\over 4}{BD}^2\\)) + \\({AC}^2\\)<\/p>\n<p>\\(\\implies\\)<strong>\u00a0 \u00a0\\({AB}^2\\) + \\({BC}^2\\) + \\({AD}^2\\) + \\({CD}^2\\) = \\({AC}^2\\) + \\({BD}^2\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Solution : We know that AD is a median of triangle ABC, then \\({AB}^2\\) + \\({AC}^2\\) = 2\\({AD}^2\\) + \\({1\\over 2}{BC}^2\\) Since the diagonals of a parallelogram bisect each other, therefore BO and DO are medians of triangles ABC and ADC respectively. \\(\\therefore\\)\u00a0 \\({AB}^2\\) + \\({BC}^2\\) = 2\\({BO}^2\\) + \\({1\\over 2}{AC}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathemerize.com\/prove-that-the-sum-of-the-squares-of-the-diagonals-of-a-parallelogram-is-equal-to-the-sum-of-the-squares-of-its-sides\/\"> <span class=\"screen-reader-text\">Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.<\/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>Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. - 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\/prove-that-the-sum-of-the-squares-of-the-diagonals-of-a-parallelogram-is-equal-to-the-sum-of-the-squares-of-its-sides\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : We know that AD is a median of triangle ABC, then ({AB}^2) + ({AC}^2) = 2({AD}^2) + ({1over 2}{BC}^2) Since the diagonals of a parallelogram bisect each other, therefore BO and DO are medians of triangles ABC and ADC respectively. (therefore)\u00a0 ({AB}^2) + ({BC}^2) = 2({BO}^2) + ({1over 2}{AC}^2)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &hellip; Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. 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(therefore)\u00a0 ({AB}^2) + ({BC}^2) = 2({BO}^2) + ({1over 2}{AC}^2)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &hellip; Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. 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