tag:blogger.com,1999:blog-6933544261975483399.post6644069377872955169..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 622: Intersecting Circles, Concyclic Points, Centers, RadiiAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-64635700124800444872015-11-29T09:12:49.141-08:002015-11-29T09:12:49.141-08:00Tr.s ACB and ADB are congruent hence < ADB = &...Tr.s ACB and ADB are congruent hence < ADB = < ACB = < AFB since AC = AF so ADFB is cyclic. Similarly ADBE is cyclic<br /><br />So AEFBD is cyclic<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11229846181787508402011-06-13T16:30:37.391-07:002011-06-13T16:30:37.391-07:00tr BEC isosceles => ang CEB= ang ECB=a
tr CAF i...tr BEC isosceles => ang CEB= ang ECB=a<br />tr CAF isosceles => ang CFA= ang ACF=a<br />thus AEFB is cyclic.<br />ACBD is a deltoid so ang D = ang C and therefore ang AEB+ ang ADB= ang AEB+ ang ACB=180 => AEBD is cyclic.<br />finally AEFBD is cyclic.Editorhttps://www.blogger.com/profile/18079120609888942700noreply@blogger.com