Question

Na Figura, M é o ponto médio do lado AC e N é o ponto Médio do lado BC. Demonstre, analiticamente, que o comprimento do segmento MN é igual à metade do comprimento do lado AB.

Answer 1

M é o ponto médio de AC, ou seja :

$\displaystyle \text M=(\frac{\text x_\text A+\text x_\text C}{2},\frac{\text y_\text A+\text y_\text C}{2}) \\\\ \text M =( \frac{0+0}{2},\frac{0+\text b}{2} )\\\\\\ \text M =(0,\frac{\text b}{2})$

N é o ponto médio de BC, ou seja :

$\displaystyle \text N=(\frac{\text x_\text B+\text x_\text C}{2},\frac{\text y_\text B+\text y_\text C}{2}) \\\\ \text N =( \frac{\text a+0}{2},\frac{0+\text b}{2} )\\\\\\ \text N =(\frac{\text a}{2},\frac{\text b}{2})$

Vamos provar que o comprimento MN é metade do comprimento AB.

$\displaystyle \text{AB}=\sqrt{(0-\text a)^2+(0-0)^2} \to \text{AB}=\sqrt{\text a^2} \\\\\ \boxed{\text{AB}=\text a\ } \\\\\\ \text{MN}=\sqrt{(0-\frac{\text a}{2})^2+(\frac{\text b}{2}-\frac{\text b}{2})^2}\to \text{MN}=\sqrt{\frac{\text a^2}{4}+0 }\\\\\\ \boxed{\text{MN}=\frac{\text a}{2}\ } \\\\\\ \underline{\text{Portanto}}: \\\\ \huge\boxed{\text{MN}=\frac{\text{AB}}{2} \ }\checkmark$