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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1270.
< EAO = < EC1O = 90-A/2 (since < C1CD = 90-C/2 and < C1DC = 90-B/2)Hence EAC1O is a cyclic quadrilateral So < AC1C = < AEO = 90Sumith PeirisMoratuwaSri Lanka
Further if AO meets DE at A1 and CA1 and C1A meet at X then O,X,F are collinear points. Further if CX cuts FD at Z and AX cuts FE at Y then XF and YZ bisect each other. (XYFZ is a parallelogram)