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Regular Pentagon and point in the minor arc

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:

pentagonSuppose 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

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