P and Q are 2 points in the area bounded by 2 rays, e and f, coming out from a point O. Describe how to construct, with a ruler and a compass only, an isosceles triangle ABC, such that his base AB is on the ray e, the point C is on the ray f, P is on AC, and Q on BC.
1. **Reflect Point Q:**
   - Construct \( Q' \), the reflection of \( Q \) in ray \( f \). This can be done by drawing a perpendicular from \( Q \) to ray \( f \) and extending it an equal distance on the other side of \( f \).

2. **Angle Calculation:**
   - Let \( \alpha \) be the angle between rays \( e \) and \( f \).
   - We need to show that \( \angle PCQ' = 180^\circ - 2\alpha \).

3. **Angle Analysis:**
   - Consider the angles formed by the points and rays:
     \[
     \angle PCQ' = \angle PCO + \angle OCQ'
     \]
   - Since \( Q' \) is the reflection of \( Q \), \( \angle OCQ' = \angle OCQ \).
   - Therefore:
     \[
     \angle PCQ' = \angle PCO + \angle OCQ
     \]
   - Note that \( \angle PCO = \angle QCP \) because \( \triangle PCO \) and \( \triangle QCO \) are congruent by reflection.
   - Thus:
     \[
     \angle PCQ' = \angle QCP + 2 \angle PCO
     \]
   - Since \( \angle QCP = \angle BCA \) and \( \angle PCO = \angle ACO \), we have:
     \[
     \angle PCQ' = \angle BCA + 2 \angle ACO
     \]
   - Using the fact that \( \angle ACO = \angle BAC - \alpha \):
     \[
     \angle PCQ' = \angle BCA + 2 (\angle BAC - \alpha)
     \]
   - Since \( \angle BAC + \angle BCA + \angle ACB = 180^\circ \) in \( \triangle ABC \), and \( \angle ACB = \alpha \):
     \[
     \angle BCA = 180^\circ - \angle BAC - \alpha
     \]
   - Substituting back:
     \[
     \angle PCQ' = (180^\circ - \angle BAC - \alpha) + 2 (\angle BAC - \alpha)
     \]
     \[
     \angle PCQ' = 180^\circ - \alpha + \angle BAC - 2\alpha
     \]
     \[
     \angle PCQ' = 180^\circ - 2\alpha
     \]

4. **Constructing Point C:**
   - Draw the circular arc \( PQ' \) with the angle \( 180^\circ - 2\alpha \) away from \( O \).
   - The intersection of this arc with ray \( f \) is point \( C \).

5. **Constructing Triangle ABC:**
   - Draw line segments \( AC \) and \( BC \).
   - Ensure that \( P \) lies on \( AC \) and \( Q \) lies on \( BC \).

The final answer is the construction of point \( \boxed{ C } \) as described, ensuring \( \triangle ABC \) is isosceles with base \( AB \) on ray \( e \), point \( C \) on ray \( f \), \( P \) on \( AC \), and \( Q \) on \( BC \).