{"id":11541,"date":"2022-07-18T14:38:02","date_gmt":"2022-07-18T09:08:02","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11541"},"modified":"2022-07-18T14:38:07","modified_gmt":"2022-07-18T09:08:07","slug":"in-the-figure-d-is-a-point-on-hypotenuse-ac-of-triangle-abc-bd-perp-ac-dm-perp-bc-and-dn-perp-ab-prove-that-i-dm2-dn-mc-ii-dn2-dm-an","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-the-figure-d-is-a-point-on-hypotenuse-ac-of-triangle-abc-bd-perp-ac-dm-perp-bc-and-dn-perp-ab-prove-that-i-dm2-dn-mc-ii-dn2-dm-an\/","title":{"rendered":"In the figure, D is a point on hypotenuse AC of \\(\\triangle\\) ABC, BD \\(\\perp\\) AC, DM \\(\\perp\\) BC and DN \\(\\perp\\) AB. Prove that (i) \\({DM}^2\\) = DN.MC (ii) \\({DN}^2\\) = DM.AN"},"content":{"rendered":"

Solution :<\/h2>\n

We have : AB \\(\\perp\\) BC and DM \\(\\perp\\) BC.\"triangle\"<\/p>\n

So,\u00a0 \u00a0 \u00a0 \u00a0 AB || DM<\/p>\n

Similarly, we have :<\/p>\n

BC \\(\\perp\\) AB and DN \\(\\perp\\) AB.<\/p>\n

So,\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 CB || DN<\/p>\n

Hence, quadrilateral BMDN is a rectangle.<\/p>\n

\\(\\therefore\\)\u00a0 BM = DN<\/p>\n

(i)<\/strong>\u00a0 In triangle BMD, we have :<\/p>\n

\\(\\angle\\)1 + \\(\\angle\\) BMD + \\(\\angle\\)2 = 180<\/p>\n

or\u00a0 \\(\\angle\\)1 + 90 + \\(\\angle\\)2 = 180<\/p>\n

\\(\\implies\\)\u00a0 \\(\\angle\\)1 + \\(\\angle\\)2 = 90<\/p>\n

Similarly, in triangle BMC, we have :<\/p>\n

\\(\\angle\\)3 + \\(\\angle\\)4 = 90<\/p>\n

Since BD \\(\\perp\\) AC, therefore<\/p>\n

\\(\\angle\\)2 + \\(\\angle\\)3 = 90<\/p>\n

Now, \\(\\angle\\)1 + \\(\\angle\\)2 = 90\u00a0 \u00a0and\u00a0 \\(\\angle\\)2 + \\(\\angle\\)3 = 90<\/p>\n

\\(\\therefore\\)\u00a0 \\(\\angle\\)1 + \\(\\angle\\)2 = \\(\\angle\\)2 + \\(\\angle\\)3<\/p>\n

So,\u00a0 \\(\\angle\\)1 = \\(\\angle\\)3<\/p>\n

Also,\u00a0 \\(\\angle\\)3 + \\(\\angle\\)4 = \\(\\angle\\)2 + \\(\\angle\\)3\u00a0 \\(\\implies\\)\u00a0 \\(\\angle\\)2 = \\(\\angle\\)4<\/p>\n

Thus, in triangles BMD and BMC, we have :<\/p>\n

\\(\\angle\\)1 = \\(\\angle\\)3\u00a0 \u00a0and\u00a0 \\(\\angle\\)2 = \\(\\angle\\)4<\/p>\n

\\(\\therefore\\) By AA similarity, we have :<\/p>\n

\\(\\triangle\\) BMD ~ \\(\\triangle\\) DMC<\/p>\n

So,\u00a0 \\(BM\\over DM\\) = \\(MD\\over MC\\)<\/p>\n

Since, BM = ND,<\/p>\n

\\(\\implies\\)\u00a0 \\(DN\\over DM\\) = \\(DM\\over MC\\)<\/p>\n

So, \\({DM}^2\\) = DN.MC<\/strong><\/p>\n

(ii)<\/strong>\u00a0 Proceeding as in (i), we can prove that<\/p>\n

\\(\\triangle\\) BND ~ \\(\\triangle\\) AND<\/p>\n

So,\u00a0 \\(BN\\over DN\\) = \\(ND\\over NA\\)<\/p>\n

Since, BN = DM,<\/p>\n

\\(\\implies\\)\u00a0 \\(DM\\over DN\\) = \\(DN\\over AN\\)<\/p>\n

So, \\({DN}^2\\) = DN.MC<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : We have : AB \\(\\perp\\) BC and DM \\(\\perp\\) BC. So,\u00a0 \u00a0 \u00a0 \u00a0 AB || DM Similarly, we have : BC \\(\\perp\\) AB and DN \\(\\perp\\) AB. So,\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 CB || DN Hence, quadrilateral BMDN is a rectangle. \\(\\therefore\\)\u00a0 BM = DN (i)\u00a0 In triangle BMD, we have : …<\/p>\n

In the figure, D is a point on hypotenuse AC of \\(\\triangle\\) ABC, BD \\(\\perp\\) AC, DM \\(\\perp\\) BC and DN \\(\\perp\\) AB. Prove that (i) \\({DM}^2\\) = DN.MC (ii) \\({DN}^2\\) = DM.AN<\/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":"\nIn the figure, D is a point on hypotenuse AC of \\(\\triangle\\) ABC, BD \\(\\perp\\) AC, DM \\(\\perp\\) BC and DN \\(\\perp\\) AB. Prove that (i) \\({DM}^2\\) = DN.MC (ii) \\({DN}^2\\) = DM.AN - 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-d-is-a-point-on-hypotenuse-ac-of-triangle-abc-bd-perp-ac-dm-perp-bc-and-dn-perp-ab-prove-that-i-dm2-dn-mc-ii-dn2-dm-an\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"In the figure, D is a point on hypotenuse AC of \\(\\triangle\\) ABC, BD \\(\\perp\\) AC, DM \\(\\perp\\) BC and DN \\(\\perp\\) AB. Prove that (i) \\({DM}^2\\) = DN.MC (ii) \\({DN}^2\\) = DM.AN - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : We have : AB (perp) BC and DM (perp) BC. So,\u00a0 \u00a0 \u00a0 \u00a0 AB || DM Similarly, we have : BC (perp) AB and DN (perp) AB. So,\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 CB || DN Hence, quadrilateral BMDN is a rectangle. (therefore)\u00a0 BM = DN (i)\u00a0 In triangle BMD, we have : … In the figure, D is a point on hypotenuse AC of (triangle) ABC, BD (perp) AC, DM (perp) BC and DN (perp) AB. 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