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Problema Proposto
42 - Dado os pontos consecutivos A, B, C e D
sobre uma reta, de modo que
AC²=AD. BD
Calcular [tex3]\frac{AB}{CD}-\frac{BD}{AC}[/tex3]

[tex3]\mathsf{
AB = x\\
BC = y\\
CD = z\\
AC^2 = AD.BD \implies (x+y)^2 = (x+y+z)(y+z) \implies\\
\boxed{x^2+xy = xz +2yz+z^2}(I)\\\frac{AB}{CD}-\frac{BD}{AC} =\frac{AB.AC-BD.CD}{AC.CD}\implies\\
\frac{x(x+y)-(y+z)z}{(x+y)z}=\frac{\underbrace{x^2+xy}-yz-z^2}{(x+y)z}\\
De(I) : \frac{xz+2yz+z^2-yz-z^2}{(x+y)z}=\frac{xz+yz}{(x+y)z}=\cancel{\frac{z(x+y)}{z(x+y)}}=\boxed{\color{red}1}

}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{
AB = x\\
BC = y\\
CD = z\\
AC^2 = AD.BD \implies (x+y)^2 = (x+y+z)(y+z) \implies\\
\boxed{x^2+xy = xz +2yz+z^2}(I)\\\frac{AB}{CD}-\frac{BD}{AC} =\frac{AB.AC-BD.CD}{AC.CD}\implies\\
\frac{x(x+y)-(y+z)z}{(x+y)z}=\frac{\underbrace{x^2+xy}-yz-z^2}{(x+y)z}\\
De(I) : \frac{xz+2yz+z^2-yz-z^2}{(x+y)z}=\frac{xz+yz}{(x+y)z}=\cancel{\frac{z(x+y)}{z(x+y)}}=\boxed{\color{red}1}

}[/tex3]