## Elementary Geometry for College Students (6th Edition)

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