tag:blogger.com,1999:blog-6933544261975483399.post8889321740464238255..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 993: Intersecting Circles, Secant, Tangent, Concyclic Points, Cyclic Quadrilateral, PerpendicularAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-56946573669671933262014-03-14T15:50:45.680-07:002014-03-14T15:50:45.680-07:00<CBD=(<CBA=<ECD)+(<DBA=<EDC)=180-&l...<CBD=(<CBA=<ECD)+(<DBA=<EDC)=180-<CED, which makes CEDB cyclic and all points CEDBF lie on a circle. Then it is obvious that <FBE=<FDE=90.<br />It also follows that <QFB=<DCB=<AOB/2=<QOB which makes points OQBF lie on a circle.<br />Ivan Bazarovnoreply@blogger.com