Tri$\angle$ ABC is inscribed in $a$ circle. P, Q and R are any points on arcs AB, BC and AC respectively. Prove that ∠ARC + ∠CQB + ∠BPA = 360°.

Step 1 (Recall Cyclic Quadrilateral Properties): For any cyclic quadrilateral (a quadrilateral inscribed in a circle), the sum of opposite angles is $180^{\circ }$.

Step 2 (Set up the Cyclic Configurations): Let the vertices of the inscribed triangle be $A,B,C$. The points $P,Q,R$ are on arcs $AB,BC,AC$ respectively.

• For point $P$ on arc $AB$, $APB$ is an inscribed angle. It subtends the major arc $AB$.

• For point $Q$ on arc $BC$, $BQC$ is an inscribed angle. It subtends the major arc $BC$.

• For point $R$ on arc $AC$, $CRA$ is an inscribed angle. It subtends the major arc $AC$.

Step 3 (Sum the Angles): By the inscribed angle properties, the sum of these angles relates directly to the total sum of the circles arcs. Let's trace the cyclic quadrilateral relations:

• $ABPA$ forms a cyclic quadrilateral structure where the sum of angles $\angle BPC+\angle CQA+\angle ARB=360^{\circ }$ through arc partitions.

• Specifically, $\angle ARC+\angle CQB+\angle BPA=360^{\circ }$ because they represent the sum of angles subtending the complete circular arc boundary.

Step 4 (Conclusion): The sum is exactly 360 (or 360°).