Given a rectangle $ABCD$, side $AB$ is longer than side $BC$. Find all the points $P$ of the side line $AB$ from which the sides $AD$ and $DC$ are seen from the point $P$ at an equal angle (i.e. $\angle APD = \angle DPC$)
1. **Define the variables and setup the problem:**
   Let \( AD = m \), \( AB = n \), and \( AP = x \). Then \( PB = n - x \). We need to find the points \( P \) on \( AB \) such that \( \angle APD = \angle DPC \).

2. **Analyze the angles:**
   Since \( \angle APD = \angle DPC \), we can use the tangent function to express these angles. We have:
   \[
   \tan \angle APD = \frac{m}{x}
   \]
   and
   \[
   \tan \angle DPC = \frac{m}{n - x}
   \]

3. **Express \(\tan \angle APC\) using the double-angle formula:**
   Since \( \angle APC = 2 \angle APD \), we use the double-angle formula for tangent:
   \[
   \tan \angle APC = \tan(2 \angle APD) = \frac{2 \tan \angle APD}{1 - \tan^2 \angle APD} = \frac{2 \cdot \frac{m}{x}}{1 - \left(\frac{m}{x}\right)^2} = \frac{2mx}{x^2 - m^2}
   \]

4. **Use the fact that \(\angle BPC + \angle APC = \pi\):**
   Since \(\angle BPC + \angle APC = \pi\), their tangent values sum to zero:
   \[
   \tan \angle BPC + \tan \angle APC = 0
   \]
   Substituting the expressions for the tangents, we get:
   \[
   \frac{m}{n - x} + \frac{2mx}{x^2 - m^2} = 0
   \]

5. **Simplify the equation:**
   Multiply through by \((n - x)(x^2 - m^2)\) to clear the denominators:
   \[
   m(x^2 - m^2) + 2mx(n - x) = 0
   \]
   Simplify and combine like terms:
   \[
   mx^2 - m^3 + 2mnx - 2mx^2 = 0
   \]
   \[
   -mx^2 + 2mnx - m^3 = 0
   \]
   Factor out \(m\):
   \[
   m(-x^2 + 2nx - m^3) = 0
   \]
   Since \(m \neq 0\), we can divide by \(m\):
   \[
   -x^2 + 2nx - m^2 = 0
   \]

6. **Solve the quadratic equation:**
   The quadratic equation is:
   \[
   x^2 - 2nx + m^2 = 0
   \]
   Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = -2n\), and \(c = m^2\):
   \[
   x = \frac{2n \pm \sqrt{(2n)^2 - 4 \cdot 1 \cdot m^2}}{2 \cdot 1} = \frac{2n \pm \sqrt{4n^2 - 4m^2}}{2} = \frac{2n \pm 2\sqrt{n^2 - m^2}}{2} = n \pm \sqrt{n^2 - m^2}
   \]

Conclusion:
The points \( P \) on \( AB \) from which the sides \( AD \) and \( DC \) are seen at equal angles are given by:
\[
x = n \pm \sqrt{n^2 - m^2}
\]

The final answer is \( \boxed{ x = n \pm \sqrt{n^2 - m^2} } \)