313. a) Let A be an arbitrary point on the circle ( A is on the arc A1A2n+1 ). Denote the side of the polygon by a, and the length of the diagonal connecting vertices through one, by b. By Ptolemy's theorem (problem II.237) for the quadrilateral AAkAk+1Ak+2 we have: ∣AAk∣a+∣AAk+2∣a=∣AAk+1∣b(k=1,2,…,2n−1). Similar relations can be written for the quadrilaterals A2nA2n+1AA1 and A2n+1AA1A2:
∣AA1∣a+∣AA2n+1∣b=∣AA2n∣a∣AA2n+1∣a+∣AA1∣b=∣AA2∣a
By adding all these equations, leaving the vertices with even numbers on the right and the odd ones on the left, we obtain the required statement.
b) Our statement follows from part a) and the result of problem I. 206. (A similar formula can be obtained in the case of internal tangency.)