{"id":11377,"date":"2022-07-08T01:09:58","date_gmt":"2022-07-07T19:39:58","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11377"},"modified":"2022-07-08T01:10:03","modified_gmt":"2022-07-07T19:40:03","slug":"sides-ab-and-ac-and-median-ad-of-a-triangle-abc-are-respectively-proportional-to-sides-pq-and-pr-and-median-pm-of-another-triangle-pqr-show-that-triangle-abc-triangle-pqr","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/sides-ab-and-ac-and-median-ad-of-a-triangle-abc-are-respectively-proportional-to-sides-pq-and-pr-and-median-pm-of-another-triangle-pqr-show-that-triangle-abc-triangle-pqr\/","title":{"rendered":"Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that \\(\\triangle\\) ABC ~ \\(\\triangle\\) PQR."},"content":{"rendered":"

Solution :<\/h2>\n

Given<\/strong> : \"triangles\"\\(\\triangle\\) ABC and \\(\\triangle\\) PQR in which AD and PM are the medians, such that<\/p>\n

\\(AB\\over PQ\\) = \\(BC\\over QR\\) = \\(AD\\over PM\\)<\/p>\n

To prove<\/strong> : \u00a0\\(\\triangle\\) ABC ~ \\(\\triangle\\) PQR<\/p>\n

Proof<\/strong> : \\(AB\\over PQ\\) = \\(BC\\over QR\\) = \\(AD\\over PM\\)<\/p>\n

\\(\\implies\\)\u00a0 \\(AB\\over PQ\\) = \\({1\\over 2}BC\\over {1\\over 2}QR\\) = \\(AD\\over PM\\)<\/p>\n

\\(\\implies\\) \\(AB\\over PQ\\) = \\(BD\\over QM\\) = \\(AD\\over PM\\)<\/p>\n

By SSS similarity,<\/p>\n

\\(\\triangle\\) ABD ~ \\(\\triangle\\) PQM<\/p>\n

\\(\\implies\\) \\(\\angle\\) B = \\(\\angle\\) Q<\/p>\n

Now, in triangle ABC and PQR, we have<\/p>\n

\\(AB\\over PQ\\) = \\(BC\\over QR\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0(given)<\/p>\n

and\u00a0 \\(\\angle\\) B = \\(\\angle\\) Q<\/p>\n

So, By SAS similarity,<\/p>\n

\\(\\triangle\\) ABC ~ \\(\\triangle\\) PQR<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Given : \\(\\triangle\\) ABC and \\(\\triangle\\) PQR in which AD and PM are the medians, such that \\(AB\\over PQ\\) = \\(BC\\over QR\\) = \\(AD\\over PM\\) To prove : \u00a0\\(\\triangle\\) ABC ~ \\(\\triangle\\) PQR Proof : \\(AB\\over PQ\\) = \\(BC\\over QR\\) = \\(AD\\over PM\\) \\(\\implies\\)\u00a0 \\(AB\\over PQ\\) = \\({1\\over 2}BC\\over {1\\over 2}QR\\) = \\(AD\\over …<\/p>\n

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that \\(\\triangle\\) ABC ~ \\(\\triangle\\) PQR.<\/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":"\nSides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that \\(\\triangle\\) ABC ~ \\(\\triangle\\) PQR. - 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\/sides-ab-and-ac-and-median-ad-of-a-triangle-abc-are-respectively-proportional-to-sides-pq-and-pr-and-median-pm-of-another-triangle-pqr-show-that-triangle-abc-triangle-pqr\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. 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