Regular Pentagon and point in the minor arc

Join Trial or Access Free Resources

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

More Posts

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

linkedin facebook pinterest youtube rss twitter instagram facebook-blank rss-blank linkedin-blank pinterest youtube twitter instagram