#### Answer

Since two sides of the triangle have the same length, $\triangle ABC$ is an isosceles triangle.

#### Work Step by Step

We can find the length of the side $\overline{AB}$:
$L_{AB} = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
$L_{AB} = \sqrt{(1-0)^2+(2-0)^2+(4-0)^2}$
$L_{AB} = \sqrt{(1)^2+(2)^2+(4)^2}$
$L_{AB} = \sqrt{1+4+16}$
$L_{AB} = \sqrt{21}$
We can find the length of the side $\overline{AC}$:
$L_{AC} = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
$L_{AC} = \sqrt{(0-0)^2+(0-0)^2+(8-0)^2}$
$L_{AC} = \sqrt{(0)^2+(0)^2+(8)^2}$
$L_{AC} = \sqrt{0+0+64}$
$L_{AC} = 8$
We can find the length of the side $\overline{BC}$:
$L_{BC} = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
$L_{BC} = \sqrt{(1-0)^2+(2-0)^2+(4-8)^2}$
$L_{BC} = \sqrt{(1)^2+(2)^2+(-4)^2}$
$L_{BC} = \sqrt{1+4+16}$
$L_{BC} = \sqrt{21}$
Since two sides of the triangle have the same length, $\triangle ABC$ is an isosceles triangle.