ABC is an equilateral triangle inscribed in a circle, and P is a point on the minor arc BC. If PB = 5 cm and PC = 8 cm, what is the length of PA?

Apply Ptolemy's theorem to the cyclic quadrilateral ABPC: PA × BC = PB × AC + PC × AB. The three sides of the equilateral triangle are equal, say s, so PA × s = 5s + 8s and PA = 5 + 8 = 13 cm. The distance to the far vertex is always the SUM of the distances to the two near ones. Treating ∠BPC as a right angle gives the trap √89 cm, but ∠BPC = 180° − 60° = 120° here, so Pythagoras does not apply; 3 cm is the difference and 6.5 cm the average.