{"id":11538,"date":"2022-07-18T14:35:34","date_gmt":"2022-07-18T09:05:34","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11538"},"modified":"2022-07-18T14:35:38","modified_gmt":"2022-07-18T09:05:38","slug":"in-the-figure-ps-is-the-bisector-of-triangle-pqr-prove-that-qsover-sr-pqover-pr","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-the-figure-ps-is-the-bisector-of-triangle-pqr-prove-that-qsover-sr-pqover-pr\/","title":{"rendered":"In the figure, PS is the bisector of \\(\\triangle\\) PQR. Prove that \\(QS\\over SR\\) = \\(PQ\\over PR\\)."},"content":{"rendered":"

Solution :<\/h2>\n

Given<\/strong> : PQR is a triangle and PS is the bisector of \\(\\angle\\) QPR meeting QR at S.\"triangle\"<\/p>\n

\\(\\therefore\\)\u00a0 \\(\\angle\\) QPS = \\(\\angle\\) SPR<\/p>\n

To Prove<\/strong> : \\(QS\\over SR\\) = \\(PQ\\over PR\\)<\/p>\n

Construction<\/strong> : Draw RT parallel to SP to cut QP produced at T.<\/p>\n

Proof<\/strong> : Since PS || TR and PR cuts them, hence we have :<\/p>\n

\\(\\angle\\) SPR = \\(\\angle\\) PRT\u00a0 \u00a0 \u00a0 …….(1)\u00a0 \u00a0 \u00a0 \u00a0(alternate angles)\"triangle\"<\/p>\n

and \\(\\angle\\) QPS = \\(\\angle\\) PTR\u00a0 \u00a0 \u00a0 \u00a0 \u00a0………(2)\u00a0 \u00a0 \u00a0(corresponding angles)<\/p>\n

but,\u00a0 \\(\\angle\\) QPS = \\(\\angle\\) PTR\u00a0 \u00a0 \u00a0 \u00a0 ………(3)\u00a0 \u00a0 \u00a0 (given)<\/p>\n

\\(\\therefore\\)\u00a0 \\(\\angle\\) PRT = \\(\\angle\\) PTR\u00a0 \u00a0 \u00a0 \u00a0[From (1) and (2)]<\/p>\n

\\(\\implies\\)\u00a0 PT = PR\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(3)\u00a0 \u00a0 \u00a0 \u00a0 (sides opposite to equal angles are equal)<\/p>\n

Now, in \\(\\triangle\\) QRT, we have :<\/p>\n

SP || RT\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(by construction)<\/p>\n

\\(\\therefore\\)\u00a0 \\(QS\\over SR\\) = \\(PQ\\over PT\\)\u00a0 \u00a0 \u00a0 \u00a0(by basic proportionality)<\/p>\n

or\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \\(QS\\over SR\\) = \\(PQ\\over PR\\)\u00a0<\/strong> \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(from 3)<\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Given : PQR is a triangle and PS is the bisector of \\(\\angle\\) QPR meeting QR at S. \\(\\therefore\\)\u00a0 \\(\\angle\\) QPS = \\(\\angle\\) SPR To Prove : \\(QS\\over SR\\) = \\(PQ\\over PR\\) Construction : Draw RT parallel to SP to cut QP produced at T. Proof : Since PS || TR and PR …<\/p>\n

In the figure, PS is the bisector of \\(\\triangle\\) PQR. Prove that \\(QS\\over SR\\) = \\(PQ\\over PR\\).<\/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, PS is the bisector of \\(\\triangle\\) PQR. Prove that \\(QS\\over SR\\) = \\(PQ\\over PR\\). - 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-ps-is-the-bisector-of-triangle-pqr-prove-that-qsover-sr-pqover-pr\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"In the figure, PS is the bisector of \\(\\triangle\\) PQR. Prove that \\(QS\\over SR\\) = \\(PQ\\over PR\\). - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : Given : PQR is a triangle and PS is the bisector of (angle) QPR meeting QR at S. (therefore)\u00a0 (angle) QPS = (angle) SPR To Prove : (QSover SR) = (PQover PR) Construction : Draw RT parallel to SP to cut QP produced at T. Proof : Since PS || TR and PR … In the figure, PS is the bisector of (triangle) PQR. 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