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Problema Proposto
36 - Dados os pontos colineares e consecuti­vos A, B. C e D se sabe que
[tex3]2 \frac{AB}{AD}=3 \frac{BC}{CD}[/tex3] e [tex3]\frac{2}{BC}-\frac{5}{AC}=1[/tex3]
Calcular CD

[tex3]\mathsf{
\frac{2}{AC-AB}-\frac{5}{AC} =1 \implies 2AC - 5AC +5AB = AC^2-AB.AC\\
AB=\frac{AC(3+AC)}{5+AC}\\
\frac{2AB}{AD} =\frac{3BC}{CD} \implies\frac{2AB}{AD} =\frac{3(AC-AB)}{CD}\implies \frac{2.\cancel{AC}(3+AC)}{AD.(5+AC)}=\frac{3\cancel{AC}(1-\frac{AC+3}{AC+5})}{CD}\implies \\\frac{AC+3}{AD} =\frac{3}{CD}\\
CD = AD - AC\implies AD = CD+AC\\
\therefore CD.AC+3CD = 3CD+3AC \implies \boxed{\color{red}CD = 3}
}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{
\frac{2}{AC-AB}-\frac{5}{AC} =1 \implies 2AC - 5AC +5AB = AC^2-AB.AC\\
AB=\frac{AC(3+AC)}{5+AC}\\
\frac{2AB}{AD} =\frac{3BC}{CD} \implies\frac{2AB}{AD} =\frac{3(AC-AB)}{CD}\implies \frac{2.\cancel{AC}(3+AC)}{AD.(5+AC)}=\frac{3\cancel{AC}(1-\frac{AC+3}{AC+5})}{CD}\implies \\\frac{AC+3}{AD} =\frac{3}{CD}\\
CD = AD - AC\implies AD = CD+AC\\
\therefore CD.AC+3CD = 3CD+3AC \implies \boxed{\color{red}CD = 3}
}[/tex3]

(Solução:EduardoRabelo)