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Problema Proposto
37 - Em uma linha reta se consideran os pontos
colineares e consecutivos A, B, C e D de
modo que:
(2x-3) . AB . CD = AD . BC
[tex3]\frac{3y+2}{AC}= \frac{3x-14}{AB} + \frac{5z-13}{AD}[/tex3]

[tex3]\mathsf{(2x-3)\frac{AB}{BC} =\frac{AD}{CD} \implies quaterna~harmômica\\
Propriedade: n\frac{AB}{BC}=m\frac{AD}{CD}\implies
Descartes: \frac{m+n}{AC}=\frac{m}{AB}+\frac{n}{AD}\\
\therefore n=2x-3; m=1\\
comparando: 3x - 14 = 1\therefore x = 5\\
3y+2 = 2x-2 \therefore y =2\\
5z-13 = 2x-3 \therefore z = 4\\
\boxed{\color{red}x=5, y=2, z = 4}



}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{(2x-3)\frac{AB}{BC} =\frac{AD}{CD} \implies quaterna~harmômica\\
Propriedade: n\frac{AB}{BC}=m\frac{AD}{CD}\implies
Descartes: \frac{m+n}{AC}=\frac{m}{AB}+\frac{n}{AD}\\
\therefore n=2x-3; m=1\\
comparando: 3x - 14 = 1\therefore x = 5\\
3y+2 = 2x-2 \therefore y =2\\
5z-13 = 2x-3 \therefore z = 4\\
\boxed{\color{red}x=5, y=2, z = 4}



}[/tex3]