Create the page "EID" on this wiki! See also the search results found.

• 2004 AMC 10B Problems/Problem 22
  ...>\angle{IDA}=\angle{IEA}=90</math> by a property of tangency, <math>\angle{EID}=90</math>, and so <math>IDAE</math> is a square. Then, since <math>IE=2</m
  5 KB (700 words) - 13:46, 6 April 2024
• 2019 AMC 8 Problems/Problem 24
  which means <math>A[\Delta AID] = 30 = A[\Delta EID]</math>.
  27 KB (4,256 words) - 19:30, 17 January 2024
• Feuerbach point
  ...0^\circ - \angle ADB = \alpha + 2 \beta - 90^\circ = \beta -\gamma, \angle EID = \angle TID \implies</cmath>
  5 KB (825 words) - 10:23, 29 December 2023