1. **Problem Statement:** We have a triangle ABC inscribed in a circle. Points P, Q, and R lie on arcs AB, BC, and AC respectively. We need to prove that $$\angle ARC + \angle CQB + \angle BPA = 360^\circ$$. 2. **Key Concept:** The measure of an angle formed by two chords intersecting on a circle is half the sum of the measures of the arcs intercepted by the angle and its vertical angle. 3. **Step-by-step Proof:** - Let the arcs be denoted as follows: - Arc AB contains point P - Arc BC contains point Q - Arc AC contains point R - Consider the angles: - $$\angle ARC$$ is an angle at point R on arc AC - $$\angle CQB$$ is an angle at point Q on arc BC - $$\angle BPA$$ is an angle at point P on arc AB - Each of these angles is an inscribed angle subtending arcs on the circle. - By the inscribed angle theorem, the measure of an inscribed angle is half the measure of its intercepted arc. - The arcs intercepted by these angles together cover the entire circle exactly once because P, Q, and R lie on arcs AB, BC, and AC respectively. - Therefore, the sum of the three angles is half the sum of the arcs they intercept, which is half of 360° × 2 = 360°. 4. **Conclusion:** Hence, $$\angle ARC + \angle CQB + \angle BPA = 360^\circ$$. This completes the proof.