{"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":"

Solution :<\/h2>\n

We know that AD is a median of triangle ABC, then\"\"<\/p>\n

\\({AB}^2\\) + \\({AC}^2\\) = 2\\({AD}^2\\) + \\({1\\over 2}{BC}^2\\)<\/p>\n

Since the diagonals of a parallelogram bisect each other, therefore BO and DO are medians of triangles ABC and ADC respectively.<\/p>\n

\\(\\therefore\\)\u00a0 \\({AB}^2\\) + \\({BC}^2\\) = 2\\({BO}^2\\) + \\({1\\over 2}{AC}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………..(1)<\/p>\n

and\u00a0 \u00a0 \u00a0 \\({AD}^2\\) + \\({CD}^2\\) = 2\\({DO}^2\\) + \\({1\\over 2}{AC}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …………(2)<\/p>\n

Adding (1) and (2), we get<\/p>\n

\\({AB}^2\\) + \\({BC}^2\\) + \\({AD}^2\\) + \\({CD}^2\\) = 2(\\({BO}^2\\) + \\({DO}^2\\)) + \\({AC}^2\\)<\/p>\n

\\(\\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

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

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 …<\/p>\n

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 »<\/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":"\nProve 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. 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