## Interesting geometry problem

Points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}, P_{6}, P_{7}$ are placed correspondingly on sides $BC, CA, AB, BC, CA, AB, BC$ of triangle $ABC$ and thease angles are equal:
$P_{1}P_{2}C=AP_{2}P_{3}=P_{3}P_{4}B=CP_{4}P_{5}=P_ {5}P_{6}A=BP_{6}P_{7}=60^{o}$
Prove that $P_{1}=P_{7}$