{"id":4063,"date":"2010-10-18T02:12:14","date_gmt":"2010-10-18T01:12:14","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=4063"},"modified":"2022-01-18T23:08:27","modified_gmt":"2022-01-18T23:08:27","slug":"dois-triangulos","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=4063","title":{"rendered":"Dois tri\u00e2ngulos"},"content":{"rendered":"<p><ul id='GTTabs_ul_4063' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_4063' class='GTTabs_curr'><a  id=\"4063_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_4063' ><a  id=\"4063_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_4063'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"4068\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=4068\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3.jpg\" data-orig-size=\"237,184\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;HP pstc4380&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}\" data-image-title=\"Tri\u00e2ngulos\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3.jpg\" class=\"alignright size-full wp-image-4068\" title=\"Tri\u00e2ngulos\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3.jpg\" alt=\"\" width=\"190\" height=\"147\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3.jpg 237w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3-150x116.jpg 150w\" sizes=\"auto, (max-width: 190px) 100vw, 190px\" \/><\/a>Na figura, tem-se:<\/p>\n<ul>\n<li>o \u00e2ngulo de v\u00e9rtice C e o \u00e2ngulo CDB s\u00e3o geometricamente iguais;<\/li>\n<li>$\\widehat{A}=40{}^\\text{o}$;<\/li>\n<li>$A\\widehat{D}B=110{}^\\text{o}$.<\/li>\n<\/ul>\n<ol>\n<li>Determina $B\\widehat{D}C$ e $D\\widehat{B}A$.<\/li>\n<li>Mostra que os tri\u00e2ngulos [ABC] e [BCD] s\u00e3o is\u00f3sceles e indica os lados iguais em cada um. Justifica convenientemente a resposta.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_4063' onClick='GTTabs_show(1,4063)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_4063'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"4068\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=4068\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3.jpg\" data-orig-size=\"237,184\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;HP pstc4380&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}\" data-image-title=\"Tri\u00e2ngulos\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3.jpg\" class=\"aligncenter size-full wp-image-4068\" title=\"Tri\u00e2ngulos\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3.jpg\" alt=\"\" width=\"237\" height=\"184\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3.jpg 237w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3-150x116.jpg 150w\" sizes=\"auto, (max-width: 237px) 100vw, 237px\" \/>\u00ad<\/a><\/p>\n<ol>\n<li>Tendo em considera\u00e7\u00e3o que os \u00e2ngulos ADB e BDC s\u00e3o suplementares, temos:\u00a0 \\[\\begin{array}{*{35}{l}}<br \/>\nB\\widehat{D}C &amp; = &amp; 180{}^\\text{o}-A\\widehat{D}B\u00a0 \\\\<br \/>\n{} &amp; = &amp; 180{}^\\text{o}-110{}^\\text{o}\u00a0 \\\\<br \/>\n{} &amp; = &amp; 70{}^\\text{o}\u00a0 \\\\<br \/>\n\\end{array}\\]Como sabemos, a soma dos tr\u00eas \u00e2ngulos internos de um tri\u00e2ngulo \u00e9 um \u00e2ngulo raso.<br \/>\nRelativamente ao tri\u00e2ngulo [ABD], vem: \\[\\begin{array}{*{35}{l}}<br \/>\nD\\widehat{B}A &amp; = &amp; 180{}^\\text{o}-(B\\widehat{A}D+A\\widehat{D}B)\u00a0 \\\\<br \/>\n{} &amp; = &amp; 180{}^\\text{o}-(40{}^\\text{o}+110{}^\\text{o})\u00a0 \\\\<br \/>\n{} &amp; = &amp; 180{}^\\text{o}-150{}^\\text{o}\u00a0 \\\\<br \/>\n{} &amp; = &amp; 30{}^\\text{o}\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\n\u00ad<\/li>\n<li>Comecemos pelo tri\u00e2ngulo [BCD].<br \/>\nOra, num tri\u00e2ngulo, a \u00e2ngulos geometricamente iguais op\u00f5em-se lados geometricamente iguais.<br \/>\nComo os \u00e2ngulos BDC e BCD s\u00e3o geometricamente iguais (dado), ent\u00e3o os lados do tri\u00e2ngulo [ACD] opostos a estes \u00e2ngulos, respetivamente, [BC] e [BD], s\u00e3o geometricamente iguais. Assim sendo, o tri\u00e2ngulo [BCD] \u00e9 is\u00f3sceles, pois possui dois lados geometricamente iguais.<\/p>\n<p>Comecemos por determinar a amplitude do \u00e2ngulo DBC: \\[\\begin{array}{*{35}{l}}<br \/>\nD\\widehat{B}C &amp; = &amp; 180{}^\\text{o}-(B\\widehat{D}C+B\\widehat{C}D)\u00a0 \\\\<br \/>\n{} &amp; = &amp; 180{}^\\text{o}-(70{}^\\text{o}+70{}^\\text{o})\u00a0 \\\\<br \/>\n{} &amp; = &amp; 180{}^\\text{o}-140{}^\\text{o}\u00a0 \\\\<br \/>\n{} &amp; = &amp; 40{}^\\text{o}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3b.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"4090\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=4090\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3b.jpg\" data-orig-size=\"237,184\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;HP pstc4380&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}\" data-image-title=\"Tri\u00e2ngulos (2)\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3b.jpg\" class=\"alignright size-full wp-image-4090\" title=\"Tri\u00e2ngulos (2)\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3b.jpg\" alt=\"\" width=\"237\" height=\"184\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3b.jpg 237w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/pag93-3b-150x116.jpg 150w\" sizes=\"auto, (max-width: 237px) 100vw, 237px\" \/><\/a>Ora, $A\\widehat{B}C=A\\widehat{B}D+D\\widehat{B}C=30{}^\\text{o}+40{}^\\text{o}=70{}^\\text{o}$.<\/p>\n<p>Ent\u00e3o, o tri\u00e2ngulo [ABC] possui dois \u00e2ngulos geometricamente iguais: os \u00e2ngulos ABC e ACB (ambos com 70\u00ba de amplitude).<br \/>\nAssim, o tri\u00e2ngulo [ABC] possui dois lados geometricamente iguais, os lados [AB] e [AC], pois, num tri\u00e2ngulo, a \u00e2ngulos geometricamente iguais op\u00f5em-se lados geometricamente iguais. Desta forma, o tri\u00e2ngulo [ABC] \u00e9 is\u00f3sceles, pois possui dois lados geometricamente iguais.<\/p>\n<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_4063' onClick='GTTabs_show(0,4063)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Na figura, tem-se: o \u00e2ngulo de v\u00e9rtice C e o \u00e2ngulo CDB s\u00e3o geometricamente iguais; $\\widehat{A}=40{}^\\text{o}$; $A\\widehat{D}B=110{}^\\text{o}$. Determina $B\\widehat{D}C$ e $D\\widehat{B}A$. Mostra que os tri\u00e2ngulos [ABC] e [BCD] s\u00e3o is\u00f3sceles e&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20618,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,102],"tags":[424,105,67,106],"series":[],"class_list":["post-4063","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-do-espaco-ao-plano","tag-8-o-ano","tag-angulos","tag-geometria","tag-triangulos"],"views":2897,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/7V2Pag093-3_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/4063","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=4063"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/4063\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20618"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4063"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4063"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4063"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=4063"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}