{"id":16147,"date":"2014-03-10T00:48:44","date_gmt":"2014-03-10T03:48:44","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=16147"},"modified":"2023-01-30T01:03:03","modified_gmt":"2023-01-30T04:03:03","slug":"teorema-do-angulo-externo-justificativas-e-algumas-aplicacoes","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/brincando-com-geometria-teorema-do-angulo-externo\/teorema-do-angulo-externo-justificativas-e-algumas-aplicacoes\/","title":{"rendered":"Teorema do \u00e2ngulo externo: justificativas e algumas aplica\u00e7\u00f5es"},"content":{"rendered":"<p style=\"margin:20px\">\n<table style=\"border:0px; height:auto; margin-left: auto; margin-right: auto; width:530px;\">\n<tr class=\"bf\">\n<td  style=\"border: 1px solid  #ffffff;\">\n<p style=\"text-align: justify; padding: 10px; border-style: dotted; border-color :#1C1C1C; border-width :thin; width:400px;background-color: #CDBA96;\"><span style=\"color: #000000; font-family: Georgia, Times, 'Times New Roman', serif; font-size: 16px;\">J\u00e1 sei, aqui vamos tentar justificar os dois resultados envolvendo \u00e2ngulo externo que observamos utilizando o GeoGebra.<\/span><\/p>\n<\/td>\n<td style=\"border: 1px solid #ffffff; \"><img decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/02\/carinha6.jpg\" style=\"border: 0px;\"  alt=\"carinha6\" width=\"80\"  class=\"alignleft size-full wp-image-3062\" \/><\/td>\n<\/tr>\n<\/table>\n<table style=\" height:auto; margin-left: auto; margin-right: auto; width:677px;\">\n<tr class=\"bf\">\n<td style=\"border: 1px solid #ffffff\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1001 alignright\" style=\"border: 0px;\" alt=\"\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" width=\"95\" height=\"115\" \/><\/td>\n<td   style=\"border: 1px solid  #ffffff;\">\n<p style=\"text-align: justify;padding: 10px; border-style: dotted; border-color :#c1c1c1; border-width :thin; width:500px;background-color: #CDBA96;\"><span style=\"color: #000000; font-family: Georgia, Times, 'Times New Roman', serif; font-size: 16px;\">Isso  mesmo!<br \/>\nMas vamos tamb\u00e9m fazer algumas aplica\u00e7\u00f5es utilizando, principalmente, o Teorema do \u00c2ngulo Externo.<\/span><\/p>\n<\/td>\n<\/tr>\n<\/table>\n<p style=\"margin:20px\">\n<div style=\"border: 8px double #8B5A2B; padding: 10px;padding-left: 20px; padding-right: 20px; text-align:justify\">\n<p style=\"margin:20px\">\n<p style=\"text-align: center; padding-right: 15px;\"><span style=\"color: #8B5A2B; font-family: verdana, Arial, Helvetica, sans-serif; font-size: 24px;\"><strong>Teorema do \u00e2ngulo externo: justificativas e algumas aplica\u00e7\u00f5es<\/strong><\/span><\/p>\n<hr style=\"height:2px;background-color:#8B5A2B;margin-top: 6px; margin-bottom: 6px;  border: 0 none\"\/>\n<hr style=\"height:2px;background-color:#8B5A2B; margin-top: 6px; margin-bottom: 6px; border: 0 none\"\/>\n<p style=\"margin:50px\">\n<div style=\"border: 3px solid #8B5A2B; padding: 5px;padding-left: 20px; padding-right: 20px; text-align: justify; color: #616161; font-family: calibri, sans-serif; font-size: 18px; line-height: 20px;\">\n<p style=\"margin:20px\">\n<p style=\"text-align: center; padding-right: 15px;\"><span style=\"color: #8B5A2B; font-family: verdana, Arial, Helvetica, sans-serif; font-size: 24px;\"><strong><em>Justificativas<\/em><\/strong><\/span><\/p>\n<hr style=\"height:2px; width:100%;background-color:#8B5A2B; margin-top: 10px; margin-bottom: 8px;border: 0 none\"\/>\n<hr style=\"height:1px; width:80%;background-color:#8B5A2B; margin-top: 8px; margin-bottom: 40px;border: 0 none\"\/>\n<div style=\"border: 2px dotted #8B5A2B; padding: 18px;\">\n <strong>Problema 1:<\/strong> Dado um tri\u00e2ngulo  [tex]ABC[\/tex], \u00e9 bastante comum denominarmos os \u00e2ngulos [tex]A\\hat{B}C[\/tex],  [tex]B\\hat{C}A[\/tex] e  [tex]C\\hat{A}B[\/tex] de \u00e2ngulos internos do tri\u00e2ngulo. Os adjacentes a esses \u00e2ngulos, obtidos mediante prolongamento dos lados do tri\u00e2ngulo, s\u00e3o denominados de \u00e2ngulos externos do tri\u00e2ngulo [tex]ABC[\/tex].<br \/>\nPor exemplo, na figura abaixo, [tex]C\\hat{B}D[\/tex] \u00e9 um \u00e2ngulo externo do tri\u00e2ngulo  [tex]ABC[\/tex], adjacente ao \u00e2ngulo interno  [tex]A\\hat{B}C[\/tex].<\/p>\n<p style=\"margin:20px\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2014\/03\/F6.png\"  style=\"border: 0px;\" alt=\"F6\" width=\"505\" height=\"270\" class=\"aligncenter size-full wp-image-16174\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2014\/03\/F6.png 505w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2014\/03\/F6-300x160.png 300w\" sizes=\"auto, (max-width: 505px) 100vw, 505px\" \/><\/p>\n<p style=\"margin:20px\">\n<p>Considerando as medidas &beta; e &theta; assinaladas na figura, justifique porque &beta; + &theta;=180&deg;.<\/span><\/p>\n<div style=\"text-align:center;\">\n<input   value=\"Justificativa\" style=\"cursor:pointer; padding: 0px; width: 200px; height: 40px; background-color:#8B5A2B; font-size: 20px; font-family: calibri, sans-serif; font-weight: bold; border: 2px solid #64411F; color: #FFFAFA;\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar a justificativa';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Justificativa';}\"    type=\"button\" \/>    <\/p>\n<div style=\"display: none;\">\n<div style=\"border: 1px dotted #8B5A2B; padding: 10px; margin-left: auto; margin-right: auto;  width: 80%; \">\n<p style=\"text-align: justify; font-family: calibri, sans-serif; font-size: 18px;  text-align: justify; color: #616161; line-height: 25px;\">\nComo [tex]C\\hat{B}D[\/tex] \u00e9 um \u00e2ngulo externo do tri\u00e2ngulo  [tex]ABC[\/tex], adjacente ao \u00e2ngulo interno  [tex]A\\hat{B}C[\/tex], os pontos [tex]A[\/tex], [tex]B[\/tex] e [tex]D[\/tex] s\u00e3o colineares, j\u00e1 que os \u00e2ngulos [tex]A\\hat{B}C[\/tex] e  [tex]C\\hat{B}D[\/tex], al\u00e9m de adjacentes, s\u00e3o suplementares.<br \/>\nPortanto, &beta; + &theta;=180&deg;. <\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p style=\"margin:30px\">\n<div style=\"border: 2px dotted #8B5A2B; padding: 18px;\">\n <strong>Um applet para recordar:<\/strong> Utilize um applet para recordar dois fatos importantes sobre congru\u00eancia de \u00e2ngulos que ser\u00e3o utilizados no <strong>Problema 2<\/strong>. \u00c9 s\u00f3 clicar no bot\u00e3o a seguir.<\/p>\n<div style=\"text-align:center;\">\n<input   value=\"Applet\" style=\"cursor:pointer; padding: 0px; width: 160px; height: 40px; background-color:#8B5A2B; font-size: 20px; font-family: calibri, sans-serif; font-weight: bold; border: 2px solid #64411F; color: #FFFAFA;\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar o applet';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Applet';}\"    type=\"button\" \/>    <\/p>\n<div style=\"display: none;\">\n<div style=\"border: 1px dotted #8B5A2B; padding: 10px; margin-left: auto; margin-right: auto;  width: 95%; \">\n<p style=\"text-align: justify; font-family: verdana, Arial, Helvetica, sans-serif; font-size: 14px; text-align: justify; color: #616161; line-height: 25px;\">\n1) Aguarde o arquivo carregar. Na janela do applet, voc\u00ea visualizar\u00e1 duas retas paralelas, [tex]p_1[\/tex] e [tex]p_2[\/tex], e dois quadradinhos.<br \/>\n2) Marque o quadradinho <span style=\"color: blue;\">\u00c2ngulos alternos internos<\/span>, clicando sobre ele. Voc\u00ea visualizar\u00e1 uma reta <span style=\"color: blue;\"> [tex]t_1[\/tex]<\/span>, transversal a [tex]p_1[\/tex] e [tex]p_2[\/tex], e dois \u00e2ngulos alternos internos com suas respectivas medidas em graus. Clique no ponto C (ou D) com qualquer bot\u00e3o do mouse, mantenha o bot\u00e3o pressionado e movimente o ponto horizontalmente. (Se voc\u00ea estiver utilizando um celular ou um tablet, toque levemente no ponto e movimente-o.)<br \/>\nO que voc\u00ea concluiu?<br \/>\n3) Desmarque o quadradinho <span style=\"color: blue;\">\u00c2ngulos alternos internos<\/span>, clicando sobre ele.<br \/>\n4) Marque o quadradinho <span style=\"color: red;\">\u00c2ngulos correspondentes<\/span>, clicando sobre ele. Voc\u00ea visualizar\u00e1 uma reta <span style=\"color: red;\"> [tex]t_2[\/tex]<\/span>, transversal a [tex]p_1[\/tex] e [tex]p_2[\/tex], e dois \u00e2ngulos correspondentes com suas respectivas medidas em graus. Clique no ponto A (ou B) com qualquer bot\u00e3o do mouse, mantenha o bot\u00e3o pressionado e movimente o ponto horizontalmente. (Se voc\u00ea estiver utilizando um celular ou um tablet, toque levemente no ponto e movimente-o.)<br \/>\nO que voc\u00ea concluiu?<br \/>\n5) Para voltar para a configura\u00e7\u00e3o inicial da janela do applet, \u00e9 s\u00f3 clicar nas setinhas circulares que aparecem no canto superior direito do applet.<\/p>\n<p style=\"text-align:center; line-height:20px;\">\n<iframe loading=\"lazy\" scrolling=\"no\" title=\"Congru\u00eancia de \u00e2ngulos\" src=\"https:\/\/www.geogebra.org\/material\/iframe\/id\/eychgen9\/width\/718\/height\/383\/border\/8B5A2B\/sfsb\/false\/smb\/false\/stb\/false\/stbh\/false\/ai\/false\/asb\/false\/sri\/true\/rc\/false\/ld\/false\/sdz\/false\/ctl\/false\" width=\"718px\" height=\"383px\"  style=\"border: 0px solid #8B5A2B;\"> <\/iframe><br \/>\n<span style=\"font-size: 14px;\">\tOBMEP_ srdg, criado com o <a href=\"http:\/\/www.geogebra.org\/\" target=\"_blank\">GeoGebra<\/a><\/span><br \/>\n<span style=\"font-family: 'Georgia'; font-size: 14px; color: #8B5A2B;text-align:center; line-height:20px;\">Observamos que o <em>applet<\/em> ajuda na visualiza\u00e7\u00e3o do resultado;<br \/>\nmas, matematicamente, n\u00e3o substitui sua demonstra\u00e7\u00e3o.<\/span><\/p>\n<\/div>\n<p style=\"margin:10px\">\n<\/div>\n<\/div>\n<\/div>\n<p style=\"margin:30px\">\n<div style=\"border: 2px dotted #8B5A2B; padding: 18px;\">\n<p style=\"text-align: justify;\"><span style=\"color: #616161; font-family: calibri, sans-serif; font-size: 18px; line-height: 20px;\"> <strong>Problema 2:<\/strong> Justifique porque a medida de um \u00e2ngulo externo de um tri\u00e2ngulo qualquer \u00e9 a soma das medidas dos \u00e2ngulos internos n\u00e3o adjacentes a esse \u00e2ngulo externo. <\/span><\/p>\n<div style=\"text-align:center;\">\n<input   value=\"Justificativa\" style=\"cursor:pointer; padding: 0px; width: 200px; height: 40px; background-color:#8B5A2B; font-size: 20px; font-family: calibri, sans-serif; font-weight: bold; border: 2px solid #64411F; color: #FFFAFA;\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar a justificativa';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Justificativa';}\"    type=\"button\" \/>    <\/p>\n<div style=\"display: none;\">\n<div style=\"border: 1px dotted #8B5A2B; padding: 10px; margin-left: auto; margin-right: auto;  width: 98%; text-align: justify; font-family: calibri, sans-serif; font-size: 18px;  text-align: justify; color: #616161; line-height: 25px;\">\nConsidere um tri\u00e2ngulo [tex]ABC[\/tex] qualquer, fixe um de seus \u00e2ngulos internos e os dois \u00e2ngulos externos adjacentes a esse \u00e2ngulo, conforme ilustra a figura a seguir.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274.png\" style=\"border: 0px;\"  alt=\"\" width=\"493\" height=\"298\" class=\"aligncenter size-full wp-image-183523\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274.png 493w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274-300x181.png 300w\" sizes=\"auto, (max-width: 493px) 100vw, 493px\" \/><br \/>\nVamos justificar a afirma\u00e7\u00e3o de que um \u00e2ngulo externo de um tri\u00e2ngulo \u00e9 a soma das medidas dos \u00e2ngulos internos n\u00e3o adjacentes a esse \u00e2ngulo externo. No caso ilustrado pela figura, vamos mostrar que [tex] \\textcolor{#FF33FF}{\\varphi}=\\textcolor{#0099FF}{\\alpha}+\\textcolor{#080}{\\beta} [\/tex].<\/p>\n<ul>\n<li><strong>Justificativa 1:<\/strong> Observe inicialmente que, conforme ilustra o applet acima, \u00e2ngulos alternos internos t\u00eam a mesma medida. Assim, podemos afirmar que, ao tra\u00e7armos uma paralela \u00e0 reta [tex]CA[\/tex] passando por [tex]B[\/tex], essa paralela define com o segmento [tex]\\overline{BA}[\/tex] um \u00e2ngulo com v\u00e9rtice em [tex]B[\/tex] cuja a medida \u00e9 [tex] \\textcolor{#0099FF}{\\alpha}[\/tex].<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274_1.png\" style=\"border: 0px;\" alt=\"\" width=\"492\" height=\"302\" class=\"aligncenter size-full wp-image-183524\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274_1.png 492w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274_1-300x184.png 300w\" sizes=\"auto, (max-width: 492px) 100vw, 492px\" \/><br \/>\nNote que a paralela \u00e0 reta [tex]CA[\/tex] passando por [tex]B[\/tex] que tra\u00e7amos define com o segmento [tex]\\overline{BC}[\/tex] um \u00e2ngulo com v\u00e9rtice em [tex]B[\/tex] cuja a medida \u00e9 [tex] \\textcolor{#0099FF}{\\alpha}+\\textcolor{#080}{\\beta}[\/tex].<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274_2.png\"  style=\"border: 0px;\"  alt=\"\" width=\"494\" height=\"302\" class=\"aligncenter size-full wp-image-183525\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274_2.png 494w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274_2-300x183.png 300w\" sizes=\"auto, (max-width: 494px) 100vw, 494px\" \/><br \/>\nAgora vamos utilizar outra propriedade ilustrada pelo applet acima: \u00e2ngulos correspondentes t\u00eam a mesma medida, e concluir que esse \u00e2ngulo de medida [tex] \\textcolor{#0099FF}{\\alpha}+\\textcolor{#080}{\\beta}[\/tex] \u00e9 congruente ao \u00e2ngulo externo  [tex]A\\hat{C}D[\/tex].<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274_3.png\"  style=\"border: 0px;\"  alt=\"\" width=\"493\" height=\"301\" class=\"aligncenter size-full wp-image-183526\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274_3.png 493w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274_3-300x183.png 300w\" sizes=\"auto, (max-width: 493px) 100vw, 493px\" \/><br \/>\nDessa forma, segue que, de fato, [tex]\\boxed{\\textcolor{#FF33FF}{\\varphi}=\\textcolor{#0099FF}{\\alpha}+\\textcolor{#080}{\\beta}}[\/tex].<br \/>\nA justificativa que fizemos, pode ser repetida para os dois outros \u00e2ngulos internos do tri\u00e2ngulo [tex]ABC[\/tex].<br \/>\n<strong>Que tal voc\u00ea tentar?<\/strong><\/li>\n<\/ul>\n<hr style=\"height:1px; width:100%;background-color:#8B5A2B; margin-top: 10px; margin-bottom: 30px;border: 0 none\"\/>\n<ul>\n<li><strong>Justificativa 2:<\/strong> Para esta justificativa, vamos utilizar uma propriedade importante da geometria:<br \/>\n<span style=\"padding-left:50px\"><strong>A soma das medidas dos \u00e2ngulos internos de um dado tri\u00e2ngulo \u00e9 [tex]180^\\circ[\/tex].<\/strong><\/span><br \/>\nAssim, observando os \u00e2ngulos internos do tri\u00e2ngulo [tex]ABC[\/tex], conclu\u00edmos que<br \/>\n[tex]\\qquad \\qquad \\textcolor{#0099FF}{\\alpha}+\\textcolor{#080}{\\beta}+\\textcolor{#996600}{\\theta}=180^\\circ[\/tex].[tex]\\qquad \\qquad \\textcolor{red}{(i)}[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274_4.png\" style=\"border: 0px;\" alt=\"\" width=\"494\" height=\"258\" class=\"aligncenter size-full wp-image-183527\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274_4.png 494w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274_4-300x157.png 300w\" sizes=\"auto, (max-width: 494px) 100vw, 494px\" \/><br \/>\nPor outro lado, o <strong>Problema 1<\/strong> nos garante que:<br \/>\n[tex]\\qquad \\qquad \\textcolor{#996600}{\\theta}+\\textcolor{#FF33FF}{\\varphi}=180^\\circ[\/tex];[tex]\\qquad \\qquad \\textcolor{red}{(ii)}[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274_4-1.png\" style=\"border: 0px;\"alt=\"\" width=\"501\" height=\"302\" class=\"aligncenter size-full wp-image-183528\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274_4-1.png 501w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2023\/01\/f274_4-1-300x181.png 300w\" sizes=\"auto, (max-width: 501px) 100vw, 501px\" \/><br \/>\nAssim,  por [tex]\\textcolor{red}{(i)}[\/tex] e [tex]\\textcolor{red}{(ii)}[\/tex], segue que:<br \/>\n[tex]\\qquad \\qquad \\textcolor{#0099FF}{\\alpha}+\\textcolor{#080}{\\beta}+\\textcolor{#996600}{\\theta}=\\textcolor{#996600}{\\theta}+\\textcolor{#FF33FF}{\\varphi}\\\\<br \/>\n\\qquad \\qquad \\textcolor{#0099FF}{\\alpha}+\\textcolor{#080}{\\beta}+\\cancel{\\textcolor{#996600}{\\theta}}=\\cancel{\\textcolor{#996600}{\\theta}}+\\textcolor{#FF33FF}{\\varphi}\\\\<br \/>\n\\qquad \\qquad \\boxed{\\textcolor{#0099FF}{\\alpha}+\\textcolor{#080}{\\beta}=\\textcolor{#FF33FF}{\\varphi}}\\\\<br \/>\n~~~~[\/tex]<br \/>\nObserva\u00e7\u00e3o: Se voc\u00ea n\u00e3o se lembra do resultado que utilizamos de que &#8220;A soma das medidas dos \u00e2ngulos internos de um dado tri\u00e2ngulo \u00e9 [tex]180^\\circ[\/tex]&#8221;, d\u00ea uma passadinha <a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=6832\" title=\"Ir para a sala de estudos.\" target=\"_blank\"><strong> nesta Sala<\/strong><\/a>. <\/span>\n<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p style=\"margin:20px\">\n<\/div>\n<p style=\"margin:40px\">\n<hr style=\"height:1px; width:30%;background-color:#8B5A2B; margin-top: 8px; margin-bottom: 8px; border: 0 none\"\/>\n<hr style=\"height:1px; width:50%;background-color:#8B5A2B; margin-top: 8px; margin-bottom: 8px;border: 0 none\"\/>\n<hr style=\"height:2px; width:70%;background-color:#8B5A2B; margin-top: 8px; margin-bottom: 8px;border: 0 none\"\/>\n<hr style=\"height:3px; width:90%;background-color:#8B5A2B; margin-top: 8px; margin-bottom: 8px; border: 0 none\"\/>\n<hr style=\"height:3px; width:90%;background-color:#8B5A2B; margin-top: 8px; margin-bottom: 8px;border: 0 none\"\/>\n<hr style=\"height:2px; width:70%;background-color:#8B5A2B; margin-top: 8px; margin-bottom: 8px;border: 0 none\"\/>\n<hr style=\"height:1px; width:50%;background-color:#8B5A2B; margin-top: 8px; margin-bottom: 8px;border: 0 none\"\/>\n<hr style=\"height:1px; width:30%;background-color:#8B5A2B; margin-top: 8px; margin-bottom: 8px;border: 0 none\"\/>\n<p style=\"margin:40px\">\n<div style=\"border: 3px solid #8B5A2B; padding: 5px;padding-left: 20px; padding-right: 20px;\">\n<p style=\"margin:20px\">\n<p style=\"text-align: center;  padding-right: 15px;\"><span style=\"color: #8B5A2B; font-family: verdana, Arial, Helvetica, sans-serif; font-size: 24px;\"><strong>Mais problemas<\/strong><\/span><\/p>\n<hr style=\"height:2px; width:100%;background-color:#8B5A2B; margin-top: 10px; margin-bottom: 8px;border: 0 none\"\/>\n<hr style=\"height:1px; width:80%;background-color:#8B5A2B; margin-top: 8px; margin-bottom: 40px;border: 0 none\"\/>\n<div style=\"border: 3px dotted #8B5A2B; padding: 5px;padding-left: 20px; padding-right: 20px;\">\n<p style=\"text-align: justify;\"><span style=\"color: #616161; font-family: calibri, sans-serif; font-size: 18px; line-height: 20px;\"> <strong>Problema 3:<\/strong> O tri\u00e2ngulo <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><em><strong>ABC<\/strong><\/em><\/span> \u00e9 is\u00f3sceles de v\u00e9rtice <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><em><strong>A<\/strong><\/em><\/span>, assim os pontos <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><em><strong>A, B<\/strong><\/em><\/span> e os pontos <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><em><strong>A,C<\/strong><\/em><\/span> determinam lados com o mesmo comprimento.<br \/>\nToma-se o ponto <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><em><strong>E<\/strong><\/em><\/span> sobre o lado [tex]\\overline{AB}[\/tex] e os pontos <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><em><strong>D<\/strong><\/em><\/span> e <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><em><strong>F<\/strong><\/em><\/span> sobre o lado [tex]\\overline{AC}[\/tex] de forma que os segmentos [tex]\\overline{BC},\\, \\overline{BD},\\, \\overline{DE},\\, \\overline{EF},\\, \\overline{FA}\\, [\/tex] tenham o mesmo comprimento.<br \/>\nDetermine, em graus, a medida &alpha; do \u00e2ngulo relativo ao v\u00e9rtice  <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><em><strong>A<\/strong><\/em><\/span>.<\/span><\/p>\n<p style=\"margin:20px\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2014\/03\/F7.png\" style=\"border: 0px;\" alt=\"F7\" width=\"246\" height=\"468\" class=\"aligncenter size-full wp-image-16198\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2014\/03\/F7.png 246w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2014\/03\/F7-157x300.png 157w\" sizes=\"auto, (max-width: 246px) 100vw, 246px\" \/><\/p>\n<p style=\"margin:20px\">\n<\/div>\n<p style=\"margin:30px\">\n<div style=\"border: 3px dotted #8B5A2B; padding: 5px;padding-left: 20px; padding-right: 20px;\">\n<p style=\"text-align: justify;\"><span style=\"color: #616161; font-family: calibri, sans-serif; font-size: 18px; line-height: 20px;\">  <strong>Problema 4:<\/strong> O tri\u00e2ngulo <span style=\" font-family: 'Cambria Math',serif;  color: #1C1C1C; font-size: 16px;\"><em><strong>ABC<\/strong><\/em><\/span> \u00e9 is\u00f3sceles de v\u00e9rtice <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><em><strong>A<\/strong><\/em><\/span>, assim os pontos <span style=\" font-family: 'Cambria Math',serif;  color: #1C1C1C; font-size: 16px;\"><em><strong>A, B<\/strong><\/em><\/span> e os pontos <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><em><strong>A,C<\/strong><\/em><\/span> determinam lados com o mesmo comprimento.<br \/>\nToma-se o ponto <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><em><strong>E<\/strong><\/em><\/span> sobre o lado [tex]\\overline{AB}[\/tex] e o ponto <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><em><strong>F<\/strong><\/em><\/span>  sobre o lado [tex]\\overline{AC}[\/tex] de forma que os \u00e2ngulos [tex]C\\hat{B}F, \\, B\\hat{C}E\\, [\/tex] tenham medidas <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><strong>60&deg;<\/strong><\/span> e <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><strong>50&deg;<\/strong><\/span>, conforme indicado na figura.<br \/>\nSabendo-se que a medida do \u00e2ngulo relativo ao v\u00e9rtice  <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><em><strong>A<\/strong><\/em><\/span> \u00e9 <span style=\" font-family: 'Cambria Math', serif;  color: #1C1C1C; font-size: 16px;\"><strong>20&deg;<\/strong><\/span>, determine, em graus, a medida &alpha; indicada na figura.<\/span><\/p>\n<p style=\"margin:20px\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2014\/03\/F8.png\" style=\"border: 0px;\" alt=\"F8\" width=\"290\" height=\"424\" class=\"aligncenter size-full wp-image-16209\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2014\/03\/F8.png 290w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2014\/03\/F8-205x300.png 205w\" sizes=\"auto, (max-width: 290px) 100vw, 290px\" \/><\/p>\n<p style=\"margin:20px\">\n<\/div>\n<p style=\"margin:20px\">\n<div style=\"border: 3px dotted #8B5A2B; padding: 5px;padding-left: 20px; padding-right: 20px;\">\n<p style=\"text-align: justify;\"><span style=\"color: #616161; font-family: calibri, sans-serif; font-size: 18px; line-height: 20px;\">  <strong>Problema 5:<\/strong> Determine, em graus, a medida &alpha; indicada na figura abaixo. <\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2014\/03\/F9.png\" style=\"border: 0px;\" alt=\"F9\" width=\"366\" height=\"363\" class=\"aligncenter size-full wp-image-16211\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2014\/03\/F9.png 366w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2014\/03\/F9-150x150.png 150w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2014\/03\/F9-300x297.png 300w\" sizes=\"auto, (max-width: 366px) 100vw, 366px\" \/><\/p>\n<p style=\"margin:20px\">\n<\/div>\n<p style=\"margin:20px\">\n<\/div>\n<table style=\"  height:auto;text-align:center;margin-left: auto; margin-right: auto; width:530px;\">\n<tr class=\"bf\">\n<td style=\"border: 1px solid #ffffff\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1001 alignright\" style=\"border: 0px;\" alt=\"\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" width=\"95\" height=\"115\" \/><\/td>\n<td  style=\"border: 1px solid  #ffffff; \">\n<p style=\"text-align: justify;padding: 10px; border-style: dotted;border-color :#1C1C1C; border-width :thin; width:400px; background-color: #CDBA96;\"><span style=\"color: #1C1C1C; font-family: Georgia, Times, 'Times New Roman', serif; font-size: 16px;\">Alguma dificuldade com os problemas 3 e 4?<br \/>\nTente um pouco mais e depois assista aos v\u00eddeos disponibilizados na pr\u00f3xima<a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=16215\" title=\"Visite a sala!\" target=\"_blank\"><span style=\"color: #FFFAFA; font-family: verdana, Arial, Helvetica, sans-serif; font-size: 14px;\"><strong> Sala <\/strong><\/span><\/a>. <\/span><\/p>\n<\/td>\n<\/tr>\n<\/table>\n<p style=\"margin:20px\">\n<hr style=\"height:2px;background-color:#8B5A2B;margin-top: 6px; margin-bottom: 6px;  border: 0 none\"\/>\n<hr style=\"height:2px;background-color:#8B5A2B; margin-top: 6px; margin-bottom: 6px; border: 0 none\"\/>\n<p style=\"text-align: right;\"><span style=\"color: #8B5A2B; font-size: 20px;\"><strong>Equipe COM &#8211; OBMEP<\/strong><\/span><\/p>\n<\/div>\n<p style=\"text-align: right;\"><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=15801\" title=\"Voltar\" target=\"_blank\"><span style=\"color: #8B5A2B;\"><strong>Voltar para a Sala de Estudos sobre o Teorema do \u00e2ngulo Externo. <\/strong><\/span><\/a><\/p>\n<p style=\"text-align: right;\"><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=16215\" title=\"Assistir ao v\u00eddeo.\" target=\"_blank\"><span style=\"color: #8B5A2B;\"><strong>Ir para a pr\u00f3xima sala. <\/strong><\/span><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>J\u00e1 sei, aqui vamos tentar justificar os dois resultados envolvendo \u00e2ngulo externo que observamos utilizando o GeoGebra. Isso mesmo! Mas vamos tamb\u00e9m fazer algumas aplica\u00e7\u00f5es utilizando, principalmente, o Teorema do \u00c2ngulo Externo. Teorema do \u00e2ngulo externo: justificativas e algumas aplica\u00e7\u00f5es Justificativas Problema 1: Dado um tri\u00e2ngulo [tex]ABC[\/tex], \u00e9 bastante comum denominarmos os \u00e2ngulos [tex]A\\hat{B}C[\/tex], [tex]B\\hat{C}A[\/tex] &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/brincando-com-geometria-teorema-do-angulo-externo\/teorema-do-angulo-externo-justificativas-e-algumas-aplicacoes\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":2,"featured_media":2970,"parent":15801,"menu_order":0,"comment_status":"open","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-16147","page","type-page","status-publish","has-post-thumbnail","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/16147","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=16147"}],"version-history":[{"count":88,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/16147\/revisions"}],"predecessor-version":[{"id":39292,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/16147\/revisions\/39292"}],"up":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/15801"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media\/2970"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=16147"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}