Consider a right -angle triangle $ABC$ with $A=90^{o}$, $AB=c$ and $AC=b$. Let $P\in AC$ and $Q\in AB$ such that $\angle APQ=\angle ABC$ and $\angle AQP = \angle ACB$. Calculate $PQ+PE+QF$, where $E$ and $F$ are the projections of $B$ and $Q$ onto $BC$, respectively.
1. **Identify the given elements and relationships:**
   - We have a right-angled triangle \( \triangle ABC \) with \( \angle A = 90^\circ \), \( AB = c \), and \( AC = b \).
   - Points \( P \in AC \) and \( Q \in AB \) are such that \( \angle APQ = \angle ABC \) and \( \angle AQP = \angle ACB \).
   - \( E \) and \( F \) are the projections of \( B \) and \( Q \) onto \( BC \), respectively.

2. **Draw the altitude \( AD \) from \( A \) to \( BC \):**
   - Let \( AD = h_a \). Since \( \triangle ABC \) is a right-angled triangle, \( h_a \) can be calculated using the formula for the altitude in a right triangle:
     \[
     h_a = \frac{bc}{\sqrt{b^2 + c^2}}
     \]

3. **Determine the midpoint \( M \) of \( PQ \):**
   - Since \( \angle APQ = \angle ABC \) and \( \angle AQP = \angle ACB \), \( \triangle APQ \) is similar to \( \triangle ABC \).
   - This similarity implies that \( P \) and \( Q \) divide \( AC \) and \( AB \) proportionally.

4. **Calculate \( PQ \):**
   - Since \( \triangle APQ \sim \triangle ABC \), the length \( PQ \) can be found using the similarity ratio. However, the exact length of \( PQ \) is not directly needed for the final calculation.

5. **Calculate \( PE \) and \( QF \):**
   - \( PE \) is the projection of \( B \) onto \( BC \), which is the length of the altitude \( AD \), so \( PE = h_a \).
   - \( QF \) is the projection of \( Q \) onto \( BC \). Since \( Q \) lies on \( AB \), and \( AB \) is perpendicular to \( BC \), \( QF \) is also equal to \( h_a \).

6. **Sum the lengths \( PQ + PE + QF \):**
   - Since \( PE = h_a \) and \( QF = h_a \), we have:
     \[
     PQ + PE + QF = PQ + h_a + h_a = PQ + 2h_a
     \]
   - Given that \( M \) is the midpoint of \( PQ \), we can express \( PQ \) in terms of \( h_a \):
     \[
     PQ = 2AM
     \]
   - Since \( AM = h_a \), we have:
     \[
     PQ = 2h_a
     \]
   - Therefore:
     \[
     PQ + PE + QF = 2h_a + 2h_a = 4h_a
     \]
   - Substituting \( h_a = \frac{bc}{\sqrt{b^2 + c^2}} \):
     \[
     PQ + PE + QF = 4 \cdot \frac{bc}{\sqrt{b^2 + c^2}} = \frac{4bc}{\sqrt{b^2 + c^2}}
     \]

The final answer is \(\boxed{\frac{4bc}{\sqrt{b^2 + c^2}}}\)