{"id":11506,"date":"2022-07-12T23:01:43","date_gmt":"2022-07-12T17:31:43","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11506"},"modified":"2022-07-12T23:03:10","modified_gmt":"2022-07-12T17:33:10","slug":"the-perpendicular-from-a-on-side-bc-of-a-triangle-abc-intersects-bc-at-d-such-that-db-3cd-see-figure-prove-that-2ab2-2ac2-bc2","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/the-perpendicular-from-a-on-side-bc-of-a-triangle-abc-intersects-bc-at-d-such-that-db-3cd-see-figure-prove-that-2ab2-2ac2-bc2\/","title":{"rendered":"The perpendicular from A on side BC of a triangle ABC intersects BC at D such that DB = 3CD (see figure). Prove that \\(2{AB}^2\\) = \\(2{AC}^2\\) + \\({BC}^2\\)"},"content":{"rendered":"

Solution :<\/h2>\n

We have : DB = 3CD\"triangle\"<\/p>\n

Now, BC = DB + CD<\/p>\n

i.e. BC = 3CD + CD\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [because BD = 3CD]<\/p>\n

BC = 4CD<\/p>\n

\\(\\therefore\\)\u00a0 \u00a0CD = \\(1\\over 4\\) BC\u00a0 and\u00a0 DB = 3CD = \\(3\\over 4\\) BC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(1)<\/p>\n

Since triangle ABD is a right triangle, right angled at D, therefore, by Pythagoras Theorem, we have :<\/p>\n

\\({AB}^2\\) = \\({AD}^2\\) + \\({DB}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(2)<\/p>\n

In triangle ACD,<\/p>\n

\\(\\angle\\) D = 90 ,\u00a0 \\({AC}^2\\) = \\({AD}^2\\) + \\({CD}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……..(3)<\/p>\n

Subtracting (3) from (2), we get<\/p>\n

\\({AB}^2\\) – \\({AC}^2\\) = \\({DB}^2\\) – \\({CD}^2\\)<\/p>\n

\\({AB}^2\\) – \\({AC}^2\\) = \\(({3\\over 4}BC)^2\\) – \\(({1\\over 4}BC)^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (using (1))<\/p>\n

\\({AB}^2\\) – \\({AC}^2\\) = \\(({9\\over 16} – {1\\over 16})\\)\\({BC}^2\\)<\/p>\n

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

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

Solution : We have : DB = 3CD Now, BC = DB + CD i.e. BC = 3CD + CD\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [because BD = 3CD] BC = 4CD \\(\\therefore\\)\u00a0 \u00a0CD = \\(1\\over 4\\) BC\u00a0 and\u00a0 DB = 3CD = \\(3\\over 4\\) BC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(1) Since triangle ABD is …<\/p>\n

The perpendicular from A on side BC of a triangle ABC intersects BC at D such that DB = 3CD (see figure). Prove that \\(2{AB}^2\\) = \\(2{AC}^2\\) + \\({BC}^2\\)<\/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":"\nThe perpendicular from A on side BC of a triangle ABC intersects BC at D such that DB = 3CD (see figure). Prove that \\(2{AB}^2\\) = \\(2{AC}^2\\) + \\({BC}^2\\) - 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\/the-perpendicular-from-a-on-side-bc-of-a-triangle-abc-intersects-bc-at-d-such-that-db-3cd-see-figure-prove-that-2ab2-2ac2-bc2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The perpendicular from A on side BC of a triangle ABC intersects BC at D such that DB = 3CD (see figure). Prove that \\(2{AB}^2\\) = \\(2{AC}^2\\) + \\({BC}^2\\) - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : We have : DB = 3CD Now, BC = DB + CD i.e. BC = 3CD + CD\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [because BD = 3CD] BC = 4CD (therefore)\u00a0 \u00a0CD = (1over 4) BC\u00a0 and\u00a0 DB = 3CD = (3over 4) BC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(1) Since triangle ABD is … The perpendicular from A on side BC of a triangle ABC intersects BC at D such that DB = 3CD (see figure). 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