**Problem: Let ABCDE be a regular pentagon inscribed in a circle. P be any point in the minor arc AE. Prove that PA + PC + PE = PB + PD**

Proof:

Suppose length of each side is ‘s’ and each diagonal is ‘x’.

Apply Ptolemy’s Theorem in PABC. We have PA . s + PC . s = PB . x

Again applying Ptolemy in PBCE, we have PB . x + PE. s = PC . x => PA. s + PC. s + PE. s = PC. x

Finally we apply Ptolemy in PBCD we have PB . s + PD. s = PC . x

Hence we have (PA + PB + PC).s = PC. x = (PB + PD).s => PA + PB + PC = PB + PD.

*Critical Idea: Ptolemy’s Theorem*

*Related*