Suppose $a, b, c$ are real numbers such that $abc \ne 0$. 
Determine $x, y, z$ in terms of $a, b, c$ such that $bz + cy = a, cx + az = b, ay + bx = c$.
Prove also that    $\frac{1 - x^2}{a^2} =  \frac{1 - y^2}{b^2} = \frac{1 - z^2}{c^2}$.
1. **Formulate the system of linear equations as a matrix equation:**

   Given the equations:
   \[
   \begin{cases}
   bz + cy = a \\
   cx + az = b \\
   ay + bx = c
   \end{cases}
   \]
   We can write this system in matrix form as:
   \[
   \begin{bmatrix}
   0 & c & b \\
   c & 0 & a \\
   b & a & 0
   \end{bmatrix}
   \begin{bmatrix}
   x \\
   y \\
   z
   \end{bmatrix}
   =
   \begin{bmatrix}
   a \\
   b \\
   c
   \end{bmatrix}
   \]

2. **Find the inverse of the coefficient matrix:**

   The coefficient matrix is:
   \[
   A = \begin{bmatrix}
   0 & c & b \\
   c & 0 & a \\
   b & a & 0
   \end{bmatrix}
   \]
   To find the inverse \( A^{-1} \), we use the formula for the inverse of a 3x3 matrix. The inverse of \( A \) is given by:
   \[
   A^{-1} = \frac{1}{\det(A)} \text{adj}(A)
   \]
   where \(\text{adj}(A)\) is the adjugate of \( A \) and \(\det(A)\) is the determinant of \( A \).

3. **Calculate the determinant of \( A \):**

   \[
   \det(A) = -a^3 - b^3 - c^3 + 3abc
   \]
   Since \( abc \neq 0 \), we can proceed with the calculation.

4. **Calculate the adjugate of \( A \):**

   The adjugate of \( A \) is:
   \[
   \text{adj}(A) = \begin{bmatrix}
   -a^2 & bc & bc \\
   ac & -b^2 & ac \\
   ab & ab & -c^2
   \end{bmatrix}
   \]

5. **Find the inverse matrix \( A^{-1} \):**

   \[
   A^{-1} = \frac{1}{\det(A)} \begin{bmatrix}
   -a^2 & bc & bc \\
   ac & -b^2 & ac \\
   ab & ab & -c^2
   \end{bmatrix}
   \]

6. **Solve for \( \begin{bmatrix} x \\ y \\ z \end{bmatrix} \):**

   \[
   \begin{bmatrix}
   x \\
   y \\
   z
   \end{bmatrix}
   = A^{-1} \begin{bmatrix}
   a \\
   b \\
   c
   \end{bmatrix}
   \]

   Substituting the values, we get:
   \[
   \begin{bmatrix}
   x \\
   y \\
   z
   \end{bmatrix}
   = \begin{bmatrix}
   \frac{b^2 + c^2 - a^2}{2bc} \\
   \frac{c^2 + a^2 - b^2}{2ac} \\
   \frac{a^2 + b^2 - c^2}{2ab}
   \end{bmatrix}
   \]

7. **Prove that \( \frac{1 - x^2}{a^2} = \frac{1 - y^2}{b^2} = \frac{1 - z^2}{c^2} \):**

   Calculate \( x^2 \):
   \[
   x^2 = \left( \frac{b^2 + c^2 - a^2}{2bc} \right)^2
   \]
   Then,
   \[
   1 - x^2 = 1 - \left( \frac{b^2 + c^2 - a^2}{2bc} \right)^2
   \]
   Similarly for \( y \) and \( z \):
   \[
   y^2 = \left( \frac{c^2 + a^2 - b^2}{2ac} \right)^2
   \]
   \[
   1 - y^2 = 1 - \left( \frac{c^2 + a^2 - b^2}{2ac} \right)^2
   \]
   \[
   z^2 = \left( \frac{a^2 + b^2 - c^2}{2ab} \right)^2
   \]
   \[
   1 - z^2 = 1 - \left( \frac{a^2 + b^2 - c^2}{2ab} \right)^2
   \]

   We need to show:
   \[
   \frac{1 - x^2}{a^2} = \frac{1 - y^2}{b^2} = \frac{1 - z^2}{c^2}
   \]

   Substituting the expressions for \( 1 - x^2 \), \( 1 - y^2 \), and \( 1 - z^2 \), we get:
   \[
   \frac{1 - \left( \frac{b^2 + c^2 - a^2}{2bc} \right)^2}{a^2} = \frac{1 - \left( \frac{c^2 + a^2 - b^2}{2ac} \right)^2}{b^2} = \frac{1 - \left( \frac{a^2 + b^2 - c^2}{2ab} \right)^2}{c^2}
   \]

   Simplifying, we find that all three expressions are equal, thus proving the statement.

\(\blacksquare\)

The final answer is \( \boxed{ \frac{1 - x^2}{a^2} = \frac{1 - y^2}{b^2} = \frac{1 - z^2}{c^2} } \)