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connect OC, <BDA1 = <COA1, so OCDA1 is on the same circle. <ODC = 90 = <OA1C =<AA1C
please explain why <BDA1 = <COA1Peter
< 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
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connect OC, <BDA1 = <COA1, so OCDA1 is on the same circle. <ODC = 90 = <OA1C =<AA1C
ReplyDeleteplease explain why <BDA1 = <COA1
DeletePeter
< DEB = 90-B/2 and < A1AE = A/2 so
ReplyDelete< EA1A = 90-B/2 - A/2 = C/2.
But < OCD = C/2 hence OCDA1 is concyclic
Therefore <OA1C = <ODC = 90
Sumith Peiris
Moratuwa
Sri Lanka
It follows that < OCA1 = B/2
ReplyDelete