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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 1269.
connect OC, <BDA1 = <COA1, so OCDA1 is on the same circle. <ODC = 90 = <OA1C =<AA1C
< DEB = 90-B/2 and < A1AE = A/2 so < EA1A = 90-B/2 - A/2 = C/2.But < OCD = C/2 hence OCDA1 is concyclicTherefore <OA1C = <ODC = 90Sumith PeirisMoratuwaSri Lanka
It follows that < OCA1 = B/2