Sabe-se que \\(\\overparen{DB}\u00a0= 3\\overparen{AC}\\) e \\(\\overparen{CB} = \\overparen{AD} = 100^\\circ \\).<\/p>\n![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/01\/9V1Pag135-6.png\")
<\/a>Considera a circunfer\u00eancia de centro O e duas retas concorrentes CD e AB cujo ponto de interse\u00e7\u00e3o \u00e9 E.<\/p>\n\n
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\n\\[\\begin{array}{*{20}{l}}{\\left\\{ {\\begin{array}{*{20}{l}}{\\overparen{DB} = 3\\overparen{AC}}\\\\{\\overparen{AC} + 100^\\circ + \\overparen{BD} + 100^\\circ = 360^\\circ }\\end{array}} \\right.}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{\\overparen{DB} = 3\\overparen{AC}}\\\\{\\overparen{AC} + 3\\overparen{AC} = 160^\\circ }\\end{array}} \\right.}& \\Leftrightarrow &{\\left\\{ {\\begin{array}{*{20}{l}}{\\overparen{AC} = 40^\\circ }\\\\{\\overparen{DB} = 120^\\circ }\\end{array}} \\right.}\\end{array}\\]
\n\u00ad<\/li>\n<\/a>Tendo em considera\u00e7\u00e3o a rela\u00e7\u00e3o entre a amplitude de um \u00e2ngulo inscrito e a do arco compreendido entre os seus lados, bem como que a soma das amplitudes dos \u00e2ngulos internos de um tri\u00e2ngulo \u00e9 igual a 180 graus, vem:
\n\\[\\begin{array}{*{20}{l}}{D\\widehat CB = \\frac{{\\overparen{BD}}}{2} = \\frac{{120^\\circ }}{2} = 60^\\circ }\\\\{A\\widehat BC = \\frac{{\\overparen{AC}}}{2} = \\frac{{40^\\circ }}{2} = 20^\\circ }\\\\{B\\widehat EC = 180^\\circ – \\left( {D\\widehat CB + A\\widehat BC} \\right) = 180^\\circ – \\left( {60^\\circ + 20^\\circ } \\right) = 100^\\circ }\\end{array}\\]<\/li>\n<\/ol>\n