Show that triangle ABC (-2, 4), B(-5, 1), and C(-6, 5) is an isosceles triangle.

To show that triangle ABC is an isosceles triangle, we need to prove that at least two sides of the triangle are equal in length.

First, let's find the lengths of the three sides of the triangle using the distance formula:

AB = √[(-5 - (-2))^2 + (1 - 4)^2]
= √[(-3)^2 + (-3)^2]
= √[9 + 9]
= √18

BC = √[(-6 - (-5))^2 + (5 - 1)^2]
= √[(-1)^2 + 4^2]
= √[1 + 16]
= √17

AC = √[(-6 - (-2))^2 + (5 - 4)^2]
= √[(-4)^2 + 1^2]
= √[16 + 1]
= √17

Now, we can see that BC = AC, so two sides of the triangle are equal in length. Therefore, triangle ABC is an isosceles triangle.

The distance between two points (x1, y1) and (x2, y2) is given by the distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Using this formula, we can find the lengths of the three sides of the triangle ABC:

$$AB = \sqrt{(-5 - (-2))^2 + (1 - 4)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$$

$$BC = \sqrt{(-6 - (-5))^2 + (5 - 1)^2} = \sqrt{(-1)^2 + (4)^2} = \sqrt{17}$$

$$AC = \sqrt{(-6 - (-2))^2 + (5 - 4)^2} = \sqrt{(-4)^2 + (1)^2} = \sqrt{17}$$

Since AB = AC, we know that triangle ABC is isosceles.