A point $P$ is given inside an acute-angled triangle $ABC$. Let $A',B',C'$ be the orthogonal projections of $P$ on sides $BC, CA, AB$ respectively. Determine the locus of points $P$ for which $\angle BAC = \angle B'A'C'$ and $\angle CBA = \angle C'B'A'$
1. **Setup and Definitions**:
   - Let \( \triangle ABC \) be an acute-angled triangle.
   - Let \( P \) be a point inside \( \triangle ABC \).
   - Let \( A', B', C' \) be the orthogonal projections of \( P \) onto sides \( BC, CA, AB \) respectively.
   - We need to determine the locus of points \( P \) such that \( \angle BAC = \angle B'A'C' \) and \( \angle CBA = \angle C'B'A' \).

2. **Complex Plane Representation**:
   - Assume \( \triangle ABC \) is inscribed in the unit circle of the complex plane.
   - Let the complex coordinates of points \( A, B, C \) be \( a, b, c \) respectively.
   - Let the complex coordinate of point \( P \) be \( p \).

3. **Orthogonal Projections**:
   - The orthogonal projection of \( P \) onto \( BC \) is given by:
     \[
     a' = \frac{1}{2}(p + b + c - \overline{p}bc)
     \]
   - The orthogonal projection of \( P \) onto \( CA \) is given by:
     \[
     b' = \frac{1}{2}(p + c + a - \overline{p}ca)
     \]
   - The orthogonal projection of \( P \) onto \( AB \) is given by:
     \[
     c' = \frac{1}{2}(p + a + b - \overline{p}ab)
     \]

4. **Similarity and Orientation**:
   - Since \( \triangle ABC \) and \( \triangle A'B'C' \) are similar and have the same orientation, we must have:
     \[
     \frac{a - b}{a' - b'} = \frac{b - c}{b' - c'}
     \]

5. **Simplifying the Condition**:
   - Substitute the expressions for \( a', b', c' \) into the similarity condition:
     \[
     \frac{a - b}{\frac{1}{2}(p + b + c - \overline{p}bc) - \frac{1}{2}(p + c + a - \overline{p}ca)} = \frac{b - c}{\frac{1}{2}(p + c + a - \overline{p}ca) - \frac{1}{2}(p + a + b - \overline{p}ab)}
     \]
   - Simplify the denominators:
     \[
     \frac{a - b}{\frac{1}{2}((b - a) + (c - c) - \overline{p}(bc - ca))} = \frac{b - c}{\frac{1}{2}((c - b) + (a - a) - \overline{p}(ca - ab))}
     \]
   - Further simplification gives:
     \[
     \frac{a - b}{\frac{1}{2}(b - a - \overline{p}(bc - ca))} = \frac{b - c}{\frac{1}{2}(c - b - \overline{p}(ca - ab))}
     \]

6. **Solving for \( \overline{p} \)**:
   - Equate the simplified expressions:
     \[
     \frac{a - b}{b - a - \overline{p}(bc - ca)} = \frac{b - c}{c - b - \overline{p}(ca - ab)}
     \]
   - Cross-multiplying and simplifying, we get:
     \[
     (a - b)(c - b - \overline{p}(ca - ab)) = (b - c)(b - a - \overline{p}(bc - ca))
     \]
   - This simplifies to:
     \[
     (a - b)(c - b) - (a - b)\overline{p}(ca - ab) = (b - c)(b - a) - (b - c)\overline{p}(bc - ca)
     \]
   - Collecting terms involving \( \overline{p} \):
     \[
     (a - b)(c - b) - (b - c)(b - a) = \overline{p}((a - b)(ca - ab) - (b - c)(bc - ca))
     \]
   - Simplifying further, we find:
     \[
     \overline{p}(a - c) = 0
     \]
   - Therefore, \( \overline{p} = 0 \) or \( a = c \).

7. **Conclusion**:
   - Since \( a = c \) is not generally true for an acute-angled triangle, we conclude that \( \overline{p} = 0 \).
   - This implies that \( P \) must be the circumcenter of \( \triangle ABC \).

\(\blacksquare\)

The final answer is the circumcenter of \( \boxed{ \triangle ABC } \).