Sejam [CD] a altura do tri\u00e2ngulo relativa \u00e0 hipotenusa, \\(x = \\overline {AD} \\) e \\(y = \\overline {DB} \\).<\/p>\n Sejam [CD] a altura do tri\u00e2ngulo relativa \u00e0 hipotenusa, \\(x = \\overline {AD} \\) e \\(y = \\overline {DB} \\).<\/p>\n<\/blockquote>\n \u00a0<\/p>\n \u00a0<\/p>\n \\[\\begin{array}{*{20}{c}}{{\\rm{S\u00e3o\\,semelhantes\\,[ACD]\\,e\\,[ABC]:}}}&{}&{{\\rm{S\u00e3o\\,semelhantes\\,[BCD]\\,e\\,[ABC]:}}}\\\\{\\frac{{\\overline {AD} }}{{\\overline {AC} }} = \\frac{{\\overline {AC} }}{{\\overline {AB} }} = \\frac{{\\overline {CD} }}{{\\overline {BC} }}}&{}&{\\frac{{\\overline {BD} }}{{\\overline {BC} }} = \\frac{{\\overline {BC} }}{{\\overline {AB} }} = \\frac{{\\overline {CD} }}{{\\overline {AC} }}}\\\\{}&{}&{}\\\\{\\frac{x}{b} = \\frac{b}{c} = \\frac{{\\overline {CD} }}{a}}&{}&{\\frac{y}{a} = \\frac{a}{c} = \\frac{{\\overline {CD} }}{b}}\\end{array}\\]<\/p>\n \u00a0<\/p>\n![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag052-T5.png\")
<\/a>Considera o tri\u00e2ngulo [ABC] ret\u00e2ngulo em C , onde \\(a = \\overline {BC} \\), \\(b = \\overline {AC} \\) e \\(c = \\overline {AB} \\).<\/p>\n\n
\n
<\/a>Considera o tri\u00e2ngulo [ABC] ret\u00e2ngulo em C , onde \\(a = \\overline {BC} \\), \\(b = \\overline {AC} \\) e \\(c = \\overline {AB} \\).<\/p>\n\n
Justifica que \\({b^2} = xc\\).<\/blockquote>\n<\/li>\n
Justifica que \\({a^2} = yc\\).<\/blockquote>\n<\/li>\n
Observando a figura e tendo em considera\u00e7\u00e3o as al\u00edneas 1. e 2., mostra que \\[{a^2} + {b^2} = {c^2}\\]<\/blockquote>\n<\/li>\n<\/ol>\n
\n
\\[\\frac{x}{b} = \\frac{b}{c}\\]
Donde resulta \\({b^2} = xc\\).
<\/li>\n
\\[\\frac{y}{a} = \\frac{a}{c}\\]
Donde resulta \\({a^2} = yc\\).
<\/li>\n
Logo, tem-se:
\\[{a^2} + {b^2} = {c^2}\\]<\/li>\n<\/ol>\n<\/a>Para esclarecimento das propor\u00e7\u00f5es consideradas acima<\/h6>\n