{"id":11369,"date":"2022-07-08T01:04:58","date_gmt":"2022-07-07T19:34:58","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11369"},"modified":"2022-07-08T01:05:03","modified_gmt":"2022-07-07T19:35:03","slug":"cd-and-gh-are-respectively-the-bisectors-of-angle-acb-and-angle-egf-such-that-d-and-h-lie-on-sides-ab-and-fe-of-triangle-abc-and-triangle-efg-respectively-if-tria","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/cd-and-gh-are-respectively-the-bisectors-of-angle-acb-and-angle-egf-such-that-d-and-h-lie-on-sides-ab-and-fe-of-triangle-abc-and-triangle-efg-respectively-if-tria\/","title":{"rendered":"CD and GH are respectively the bisectors of \\(\\angle\\) ACB and \\(\\angle\\) EGF such that D and H lie on sides AB and FE of \\(\\triangle\\) ABC and \\(\\triangle\\) EFG respectively. If \\(\\triangle\\) ABC ~ \\(\\triangle\\) FEG show that (i) \\(CD\\over GH\\) = \\(AC\\over FG\\) (ii) \\(\\triangle\\) DCB ~ \\(\\triangle\\) HGE (iii) \\(\\triangle\\) DCA ~ \\(\\triangle\\) HGF"},"content":{"rendered":"

Solution :<\/h2>\n

Given<\/strong> : \"triangles\"\\(\\triangle\\) ABC ~ \\(\\triangle\\)\u00a0 FEG and CD and GH are bisectors of \\(\\angle\\) C and \\(\\angle\\) G respectively.<\/p>\n

(i)<\/strong> In triangle ACD and FGH\u00a0<\/p>\n

\\(\\angle\\) A = \\(\\angle\\) F\u00a0 \u00a0 \u00a0 \u00a0 \u00a0(\\(\\triangle\\) ABC ~ \\(\\triangle\\)\u00a0 FEG)<\/p>\n

Since it is given that CD bisects \\(\\angle\\) C and GH bisects \\(\\angle\\) G and \\(\\angle\\) C = \\(\\angle\\) G,<\/p>\n

\\(\\implies\\)\u00a0 \u00a0\\(\\angle\\) ACD = \\(\\angle\\) FGH<\/p>\n

\\(\\therefore\\)\u00a0 By AA similarity,<\/p>\n

\\(\\triangle\\) ACD ~ \\(\\triangle\\) FGH<\/p>\n

\\(\\implies\\) \\(CD\\over GH\\) = \\(AC\\over FG\\)<\/strong><\/p>\n

(ii)<\/strong>\u00a0 In triangle DCB and HGE,<\/p>\n

\\(\\angle\\) B = \\(\\angle\\) E\u00a0 \u00a0 \u00a0 \u00a0 \u00a0(\\(\\triangle\\) ABC ~ \\(\\triangle\\)\u00a0 FGH)<\/p>\n

Since it is given that CD bisects \\(\\angle\\) C and GH bisects \\(\\angle\\) G and \\(\\angle\\) C = \\(\\angle\\) G,<\/p>\n

\\(\\implies\\)\u00a0 \u00a0\\(\\angle\\) DCB = \\(\\angle\\) HGE<\/p>\n

\\(\\therefore\\)\u00a0 By AA similarity,<\/p>\n

\\(\\triangle\\) DCB ~ \\(\\triangle\\) HGE<\/strong><\/p>\n

(iii)<\/strong>\u00a0In triangle DCA and HGF<\/p>\n

\\(\\angle\\) A = \\(\\angle\\) F\u00a0 \u00a0 \u00a0 \u00a0 \u00a0(\\(\\triangle\\) ABC ~ \\(\\triangle\\)\u00a0 FEG)<\/p>\n

Since it is given that CD bisects \\(\\angle\\) C and GH bisects \\(\\angle\\) G and \\(\\angle\\) C = \\(\\angle\\) G,<\/p>\n

\\(\\implies\\)\u00a0 \u00a0\\(\\angle\\) DCA = \\(\\angle\\) HGF<\/p>\n

\\(\\therefore\\)\u00a0 By AA similarity,<\/p>\n

\\(\\triangle\\) DCA ~ \\(\\triangle\\) HGF<\/strong><\/p>\n\n\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Given : \\(\\triangle\\) ABC ~ \\(\\triangle\\)\u00a0 FEG and CD and GH are bisectors of \\(\\angle\\) C and \\(\\angle\\) G respectively. (i) In triangle ACD and FGH\u00a0 \\(\\angle\\) A = \\(\\angle\\) F\u00a0 \u00a0 \u00a0 \u00a0 \u00a0(\\(\\triangle\\) ABC ~ \\(\\triangle\\)\u00a0 FEG) Since it is given that CD bisects \\(\\angle\\) C and GH bisects \\(\\angle\\) G …<\/p>\n

CD and GH are respectively the bisectors of \\(\\angle\\) ACB and \\(\\angle\\) EGF such that D and H lie on sides AB and FE of \\(\\triangle\\) ABC and \\(\\triangle\\) EFG respectively. If \\(\\triangle\\) ABC ~ \\(\\triangle\\) FEG show that (i) \\(CD\\over GH\\) = \\(AC\\over FG\\) (ii) \\(\\triangle\\) DCB ~ \\(\\triangle\\) HGE (iii) \\(\\triangle\\) DCA ~ \\(\\triangle\\) HGF<\/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":"\nCD and GH are respectively the bisectors of \\(\\angle\\) ACB and \\(\\angle\\) EGF such that D and H lie on sides AB and FE of \\(\\triangle\\) ABC and \\(\\triangle\\) EFG respectively. If \\(\\triangle\\) ABC ~ \\(\\triangle\\) FEG show that (i) \\(CD\\over GH\\) = \\(AC\\over FG\\) (ii) \\(\\triangle\\) DCB ~ \\(\\triangle\\) HGE (iii) \\(\\triangle\\) DCA ~ \\(\\triangle\\) HGF - 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\/cd-and-gh-are-respectively-the-bisectors-of-angle-acb-and-angle-egf-such-that-d-and-h-lie-on-sides-ab-and-fe-of-triangle-abc-and-triangle-efg-respectively-if-tria\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"CD and GH are respectively the bisectors of \\(\\angle\\) ACB and \\(\\angle\\) EGF such that D and H lie on sides AB and FE of \\(\\triangle\\) ABC and \\(\\triangle\\) EFG respectively. If \\(\\triangle\\) ABC ~ \\(\\triangle\\) FEG show that (i) \\(CD\\over GH\\) = \\(AC\\over FG\\) (ii) \\(\\triangle\\) DCB ~ \\(\\triangle\\) HGE (iii) \\(\\triangle\\) DCA ~ \\(\\triangle\\) HGF - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : Given : (triangle) ABC ~ (triangle)\u00a0 FEG and CD and GH are bisectors of (angle) C and (angle) G respectively. (i) In triangle ACD and FGH\u00a0 (angle) A = (angle) F\u00a0 \u00a0 \u00a0 \u00a0 \u00a0((triangle) ABC ~ (triangle)\u00a0 FEG) Since it is given that CD bisects (angle) C and GH bisects (angle) G … CD and GH are respectively the bisectors of (angle) ACB and (angle) EGF such that D and H lie on sides AB and FE of (triangle) ABC and (triangle) EFG respectively. If (triangle) ABC ~ (triangle) FEG show that (i) (CDover GH) = (ACover FG) (ii) (triangle) DCB ~ (triangle) HGE (iii) (triangle) DCA ~ (triangle) HGF Read More »\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mathemerize.com\/cd-and-gh-are-respectively-the-bisectors-of-angle-acb-and-angle-egf-such-that-d-and-h-lie-on-sides-ab-and-fe-of-triangle-abc-and-triangle-efg-respectively-if-tria\/\" \/>\n<meta property=\"og:site_name\" content=\"Mathemerize\" \/>\n<meta property=\"article:published_time\" content=\"2022-07-07T19:34:58+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2022-07-07T19:35:03+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-3-10-300x127.jpeg\" \/>\n<meta name=\"author\" content=\"mathemerize\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"mathemerize\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/mathemerize.com\/cd-and-gh-are-respectively-the-bisectors-of-angle-acb-and-angle-egf-such-that-d-and-h-lie-on-sides-ab-and-fe-of-triangle-abc-and-triangle-efg-respectively-if-tria\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/mathemerize.com\/cd-and-gh-are-respectively-the-bisectors-of-angle-acb-and-angle-egf-such-that-d-and-h-lie-on-sides-ab-and-fe-of-triangle-abc-and-triangle-efg-respectively-if-tria\/\"},\"author\":{\"name\":\"mathemerize\",\"@id\":\"https:\/\/mathemerize.com\/#\/schema\/person\/104c8bc54f90618130a6665299bc55df\"},\"headline\":\"CD and GH are respectively the bisectors of \\\\(\\\\angle\\\\) ACB and \\\\(\\\\angle\\\\) EGF such that D and H lie on sides AB and FE of \\\\(\\\\triangle\\\\) ABC and \\\\(\\\\triangle\\\\) EFG respectively. 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