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  • ...>\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
    7 KB (1,024 words) - 17:27, 19 July 2024
  • which means <math>A[\Delta AID] = 30 = A[\Delta EID]</math>.
    27 KB (4,265 words) - 00:25, 9 December 2024
  • ...0^\circ - \angle ADB = \alpha + 2 \beta - 90^\circ = \beta -\gamma, \angle EID = \angle TID \implies</cmath>
    6 KB (1,046 words) - 13:39, 12 December 2024
  • ...ath>EH</math> intersects <math>AD</math> at <math>J.</math> If angle <math>EID=60^{\circ},</math> the area of quadrilateral <math>EIDJ</math> is
    10 KB (1,535 words) - 17:44, 20 October 2024
  • ...ath>EH</math> intersects <math>AD</math> at <math>J.</math> If angle <math>EID=60^{\circ},</math> the area of quadrilateral <math>EIDJ</math> is ...90^{\circ}</math> because they are angles of the squares, and <math>\angle EID=60^{\circ}</math> given by the question. Similar to solution 1, if we draw
    3 KB (534 words) - 18:16, 18 July 2024