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• 2000 JBMO Problems/Problem 3
  Thus, we have <math>AK = AP</math>, and so <math>\angle AKP = \angle KPA = \alpha</math> So in <math>\triangle EKG, \angle KGE = 180 - (\angle AKP + \angle PEF)</math>
  3 KB (481 words) - 09:18, 11 October 2024