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 inscribed angle in a circle is half the measure of its intercepted arc. 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 - By the inscribed angle theorem: $$\angle ARC = \frac{1}{2} \times \text{arc} AB$$ $$\angle CQB = \frac{1}{2} \times \text{arc} BC$$ $$\angle BPA = \frac{1}{2} \times \text{arc} CA$$ - Summing these angles: $$\angle ARC + \angle CQB + \angle BPA = \frac{1}{2} (\text{arc} AB + \text{arc} BC + \text{arc} CA)$$ - Since arcs AB, BC, and CA together form the entire circle, their sum is: $$\text{arc} AB + \text{arc} BC + \text{arc} CA = 360^\circ$$ - Therefore: $$\angle ARC + \angle CQB + \angle BPA = \frac{1}{2} \times 360^\circ = 180^\circ$$ 4. **Re-examining the problem:** The problem states the sum is 360 degrees, but the sum of these inscribed angles equals 180 degrees by the inscribed angle theorem. 5. **Alternative interpretation:** If the angles are taken as reflex angles (the larger angle around the circle), then each angle is $$360^\circ - \text{inscribed angle}$$. - Then the sum is: $$\sum (360^\circ - \text{inscribed angle}) = 3 \times 360^\circ - 180^\circ = 1080^\circ - 180^\circ = 900^\circ$$ This does not match 360 degrees either. 6. **Conclusion:** The sum of the three inscribed angles as defined is $$180^\circ$$, not $$360^\circ$$. If the problem intends the sum of the arcs intercepted by these angles, then: $$\text{arc} ARC + \text{arc} CQB + \text{arc} BPA = 360^\circ$$ which is true since these arcs cover the entire circle. **Final answer:** $$\boxed{\angle ARC + \angle CQB + \angle BPA = 180^\circ}$$