tag:blogger.com,1999:blog-6933544261975483399.post8162162614417359335..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 938: Cyclic Quadrilateral, Triangle, Circumcircle, Circles, Circumcenter, Circumradius, Distance, Metric RelationsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-50609402428151625272013-12-05T17:49:01.999-08:002013-12-05T17:49:01.999-08:00Let <BAE=a and <EBA=b. Then <OO1E=1/2<...Let <BAE=a and <EBA=b. Then <OO1E=1/2<BO1A-<BO1E=(a+b)-2a=b-a. From cyclic condition we find that <EO2O=1/2<CO2D-<CO2E=b-a as well. If AB and CD meet at point P, <DPB=<DCA-<BAE=b-a. <O1OO2 is supplementary to <DPB because OO1 and OO2 perpendicular to AB and CD respective. That makes EO2OO1 a parallelogram. Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-51911860044799011772013-12-05T03:29:03.410-08:002013-12-05T03:29:03.410-08:00Conjecture: O₁OO₂E form a parallelogram.
If so, ...Conjecture: O₁OO₂E form a parallelogram. <br /><br />If so, then using parallelogram law, <br />EO² + 9² = 2(4² + 6²)<br />EO = √23<br /><br />Unfortunately, I still need some time to prove / disprove my assertion. Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.com