The problem involves proving that the triangle $\triangle ABC$, with vertices $A(0, 0)$, $B(3, 8)$, and $C(6, 0)$, is isosceles and finding the exact lengths of its sides.

Step 1: Find the lengths of the sides using the distance formula.

The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$

1. Length of $AB$:

Using $A(0, 0)$ and $B(3, 8)$: $AB = \sqrt{(3 - 0)^2 + (8 - 0)^2} = \sqrt{3^2 + 8^2} = \sqrt{9 + 64} = \sqrt{73}.$

2. Length of $BC$:

Using $B(3, 8)$ and $C(6, 0)$: $BC = \sqrt{(6 - 3)^2 + (0 - 8)^2} = \sqrt{3^2 + (-8)^2} = \sqrt{9 + 64} = \sqrt{73}.$

3. Length of $AC$:

Using $A(0, 0)$ and $C(6, 0)$: $AC = \sqrt{(6 - 0)^2 + (0 - 0)^2} = \sqrt{6^2 + 0^2} = \sqrt{36} = 6.$

Step 2: Prove $\triangle ABC$ is isosceles.

An isosceles triangle has at least two equal sides. From the above calculations:

• $AB = \sqrt{73}$,
• $BC = \sqrt{73}$,
• $AC = 6$.

Since $AB = BC$, $\triangle ABC$ is isosceles.

Final Answer:

1. The exact lengths of the sides are:
   • $AB = \sqrt{73}$,
   • $BC = \sqrt{73}$,
   • $AC = 6$.
2. $\triangle ABC$ is isosceles because $AB = BC$.

Would you like a deeper explanation of the distance formula, or assistance visualizing this triangle on a coordinate plane?

Related Questions:

1. What is the perimeter of $\triangle ABC$ in millimeters?
2. How can you calculate the area of $\triangle ABC$ using its coordinates?
3. Can this isosceles triangle ever be classified as equilateral? Why or why not?
4. What are the slopes of the sides $AB$, $BC$, and $AC$?
5. How does the coordinate system influence the physical dimensions of the triangle?

Tip:

When proving a triangle is isosceles, always compare the exact lengths of its sides carefully using the distance formula!