Estude! Com Prof. Caju

Problema Proposto
38 - Sobre uma linha reta se consideram os pontos consecutivos A, B, C e D de modo que formam uma quaterna harmônica.

Se: [tex3]\frac{a}{AC} + \frac{b}{CD} = \frac{c}{BD} + \frac{d}{AB}[/tex3]

Calcular: [tex3]a + b + c + d[/tex3]

[tex3]\mathsf{
Quaterna ~Harmônica: \frac{AB}{BC}=\frac{AD}{CD}\implies \frac{AB}{AC-AB}=\frac{AD}{AD-AC}\\
\frac{AC-AB}{AB}=\frac{AD-AC}{AD}\implies \frac{AC}{AB}-1=1-\frac{AC}{AD}\\
\therefore \frac{AC}{AB}+\frac{AC}{AD}=2(\div AC)\implies \frac{1}{AB}+\frac{1}{AD}=\frac{2}{AC}(I)\\
Analogamente: \frac{AD-BD}{BD-CD}=\frac{AD}{CD} \implies \frac{AD-BD}{AD}=\frac{BD-CD}{CD}\\
1-\frac{BD}{AD}=\frac{BD}{CD}-1\therefore \frac{BD}{CD}+\frac{BD}{AD}=2(\div BD)\\
\frac{1}{CD}+\frac{1}{AD}=\frac{2}{BD}(II)\\
De(I) e (II) \frac{2}{AC}+\frac{1}{CD}=\frac{2}{BD}+\frac{1}{AB}\\
Comparando> : \boxed{\color{red}a+b+c+d = 2+1+2+1 = 6}
}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{
Quaterna ~Harmônica: \frac{AB}{BC}=\frac{AD}{CD}\implies \frac{AB}{AC-AB}=\frac{AD}{AD-AC}\\
\frac{AC-AB}{AB}=\frac{AD-AC}{AD}\implies \frac{AC}{AB}-1=1-\frac{AC}{AD}\\
\therefore \frac{AC}{AB}+\frac{AC}{AD}=2(\div AC)\implies \frac{1}{AB}+\frac{1}{AD}=\frac{2}{AC}(I)\\
Analogamente: \frac{AD-BD}{BD-CD}=\frac{AD}{CD} \implies \frac{AD-BD}{AD}=\frac{BD-CD}{CD}\\
1-\frac{BD}{AD}=\frac{BD}{CD}-1\therefore \frac{BD}{CD}+\frac{BD}{AD}=2(\div BD)\\
\frac{1}{CD}+\frac{1}{AD}=\frac{2}{BD}(II)\\
De(I) e (II) \frac{2}{AC}+\frac{1}{CD}=\frac{2}{BD}+\frac{1}{AB}\\
Comparando> : \boxed{\color{red}a+b+c+d = 2+1+2+1 = 6}
}[/tex3]

(Solução: encontrada por geobson)