Let $f$ be a linear transformation defined by a matrix $A=\left(
\begin{array}{cc}
a & b \\
c & d
\end{array}
\right)$. For any two points $P,\ Q$, which are not different from the origin, such that $\frac{OP'}{OP}=\frac{OQ'}{OQ}$ , let $P',\ Q'$ be the images of $P,\ Q$ by $f$ respectively.

(1) Prove that $a^2+c^2=b^2+d^2$.

(2) Find $A$ such that the point $(1,\ \sqrt{3})$ is mapped onto the point $(-4,\ 0)$ by the linear transformation $f$.

50 points
### Part (1): Prove that \(a^2 + c^2 = b^2 + d^2\)

1. **Define the transformation and points:**
   Let \(P = (x_1, y_1)\) and \(Q = (x_2, y_2)\) be two points in the plane. The images of these points under the linear transformation \(f\) defined by the matrix \(A\) are:
   \[
   P' = A \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = \begin{pmatrix} ax_1 + by_1 \\ cx_1 + dy_1 \end{pmatrix}
   \]
   \[
   Q' = A \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} ax_2 + by_2 \\ cx_2 + dy_2 \end{pmatrix}
   \]

2. **Given condition:**
   The problem states that \(\frac{OP'}{OP} = \frac{OQ'}{OQ}\), where \(O\) is the origin. This implies that the ratio of the magnitudes of the images to the magnitudes of the original points is the same:
   \[
   \frac{\|P'\|}{\|P\|} = \frac{\|Q'\|}{\|Q\|}
   \]

3. **Express the magnitudes:**
   The magnitudes of \(P\) and \(P'\) are:
   \[
   \|P\| = \sqrt{x_1^2 + y_1^2}
   \]
   \[
   \|P'\| = \sqrt{(ax_1 + by_1)^2 + (cx_1 + dy_1)^2}
   \]

   Similarly, for \(Q\) and \(Q'\):
   \[
   \|Q\| = \sqrt{x_2^2 + y_2^2}
   \]
   \[
   \|Q'\| = \sqrt{(ax_2 + by_2)^2 + (cx_2 + dy_2)^2}
   \]

4. **Apply the given condition:**
   \[
   \frac{\sqrt{(ax_1 + by_1)^2 + (cx_1 + dy_1)^2}}{\sqrt{x_1^2 + y_1^2}} = \frac{\sqrt{(ax_2 + by_2)^2 + (cx_2 + dy_2)^2}}{\sqrt{x_2^2 + y_2^2}}
   \]

5. **Square both sides to eliminate the square roots:**
   \[
   \frac{(ax_1 + by_1)^2 + (cx_1 + dy_1)^2}{x_1^2 + y_1^2} = \frac{(ax_2 + by_2)^2 + (cx_2 + dy_2)^2}{x_2^2 + y_2^2}
   \]

6. **Simplify the expressions:**
   \[
   \frac{a^2x_1^2 + 2abx_1y_1 + b^2y_1^2 + c^2x_1^2 + 2cdx_1y_1 + d^2y_1^2}{x_1^2 + y_1^2} = \frac{a^2x_2^2 + 2abx_2y_2 + b^2y_2^2 + c^2x_2^2 + 2cdx_2y_2 + d^2y_2^2}{x_2^2 + y_2^2}
   \]

7. **Combine like terms:**
   \[
   \frac{(a^2 + c^2)x_1^2 + (b^2 + d^2)y_1^2 + 2(ab + cd)x_1y_1}{x_1^2 + y_1^2} = \frac{(a^2 + c^2)x_2^2 + (b^2 + d^2)y_2^2 + 2(ab + cd)x_2y_2}{x_2^2 + y_2^2}
   \]

8. **Since \(P\) and \(Q\) are arbitrary points, the coefficients of \(x_1^2\), \(y_1^2\), and \(x_1y_1\) must be equal to the coefficients of \(x_2^2\), \(y_2^2\), and \(x_2y_2\) respectively:**
   \[
   a^2 + c^2 = b^2 + d^2
   \]

Thus, we have proven that \(a^2 + c^2 = b^2 + d^2\).

### Part (2): Find \(A\) such that \((1, \sqrt{3})\) is mapped onto \((-4, 0)\)

1. **Set up the transformation equation:**
   \[
   A \begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix} = \begin{pmatrix} -4 \\ 0 \end{pmatrix}
   \]

2. **Express the transformation in terms of the matrix elements:**
   \[
   \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix} = \begin{pmatrix} a + b\sqrt{3} \\ c + d\sqrt{3} \end{pmatrix} = \begin{pmatrix} -4 \\ 0 \end{pmatrix}
   \]

3. **Set up the system of equations:**
   \[
   a + b\sqrt{3} = -4
   \]
   \[
   c + d\sqrt{3} = 0
   \]

4. **Solve for \(a\) and \(b\):**
   \[
   a + b\sqrt{3} = -4 \implies a = -4 - b\sqrt{3}
   \]

5. **Solve for \(c\) and \(d\):**
   \[
   c + d\sqrt{3} = 0 \implies c = -d\sqrt{3}
   \]

6. **Choose specific values for \(b\) and \(d\) to satisfy the condition \(a^2 + c^2 = b^2 + d^2\):**
   Let \(b = -\sqrt{3}\) and \(d = 1\):
   \[
   a = -4 - (-\sqrt{3})\sqrt{3} = -4 + 3 = -1
   \]
   \[
   c = -d\sqrt{3} = -1\sqrt{3} = -\sqrt{3}
   \]

7. **Verify the condition \(a^2 + c^2 = b^2 + d^2\):**
   \[
   a^2 + c^2 = (-1)^2 + (-\sqrt{3})^2 = 1 + 3 = 4
   \]
   \[
   b^2 + d^2 = (-\sqrt{3})^2 + 1^2 = 3 + 1 = 4
   \]

Thus, the matrix \(A\) that maps \((1, \sqrt{3})\) to \((-4, 0)\) is:
\[
A = \begin{pmatrix} -1 & -\sqrt{3} \\ -\sqrt{3} & 1 \end{pmatrix}
\]

The final answer is \( \boxed{ A = \begin{pmatrix} -1 & -\sqrt{3} \\ -\sqrt{3} & 1 \end{pmatrix} } \)