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• 2003 AIME I Problems/Problem 10
  ...- 60^\circ - 150^\circ = 150^\circ</math>. We now see that <math>\triangle MNB</math> and <math>\triangle CNB</math> are congruent. Therefore, <math>CB =
  7 KB (1,058 words) - 01:41, 6 December 2022
• Pascal's Theorem
  <math> \angle CAB \equiv \angle MAB \equiv 180^\circ -\angle MNB \equiv \angle MND \equiv 180^\circ - \angle MCD \equiv 180^\circ - \angle A
  5 KB (833 words) - 15:23, 6 July 2024
• 2009 AIME II Problems/Problem 15
  ...= \tan\left(\frac{\pi}{4} - a\right) = \tan((\widehat{MB})/2) = \tan\angle{MNB} = \frac{3}{4}</cmath>,
  11 KB (1,862 words) - 21:23, 23 May 2024
• 2014 IMO Problems/Problem 4
  ...CMN}=\angle{AMN}=\angle{AQP}=\angle{A}</math>, so <math>\angle{CMN}+\angle{MNB}=A</math>, which gives that <math>\angle{BKC}=180-\angle{A}</math> and we a
  13 KB (1,811 words) - 14:36, 1 June 2024
• 2015 AIME I Problems/Problem 4
  <math>\triangle MNB</math> is equilateral triangle.
  5 KB (852 words) - 00:12, 8 May 2024
• 1965 AHSME Problems/Problem 38
  mnb&=b+c+mc \\
  3 KB (495 words) - 11:39, 20 July 2024