To solve the problem:

Given:

1. $AB = AC = CD$
2. $BC = 8$
3. $\triangle ABC$ is isosceles with $\angle BCA = 90^\circ$.

We need to find the area of $\triangle BCD$.

Step 1: Analyze the triangle $\triangle ABC$

• Since $AB = AC$, $\triangle ABC$ is an isosceles right triangle.
• The relationship between the sides is: $AB = AC = \sqrt{BC^2 / 2} = \sqrt{8^2 / 2} = \sqrt{32} = 4\sqrt{2}.$

Step 2: Determine $CD$

• From the problem, $CD = AB = AC = 4\sqrt{2}$.

Step 3: Find the area of $\triangle BCD$

1. $\triangle BCD$ is a right triangle with:

   • $BC = 8$ (base),
   • $CD = 4\sqrt{2}$ (height).

2. Area of $\triangle BCD$: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 4\sqrt{2} = 16\sqrt{2}.$

Final Answer:

The area of $\triangle BCD$ is $16\sqrt{2}$.

Would you like further details or clarification? Here are five related questions to deepen understanding:

1. How can the Pythagorean theorem confirm the side lengths in $\triangle ABC$?
2. What is the geometric significance of the isosceles right triangle in this context?
3. How can $\triangle BCD$'s area be verified using coordinate geometry?
4. What happens to the area if $CD$ changes but $AB = AC$?
5. Can you calculate the perimeter of $\triangle BCD$?

Tip: When solving problems with isosceles right triangles, leverage the $1:1:\sqrt{2}$ ratio for faster computations.