tag:blogger.com,1999:blog-6933544261975483399.post230575119749012624..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1365: Circle, Tangent, Secant, Chord, Parallel, Midpoint, Sketch, iPad AppsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-50393694050373788532018-10-08T20:44:01.707-07:002018-10-08T20:44:01.707-07:00in quadrilateral MOAB , we have ∠ (BMA)= ∠ (BOA)
s...in quadrilateral MOAB , we have ∠ (BMA)= ∠ (BOA)<br />so MOAB is cyclicPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68408230236801620292018-10-07T07:03:43.095-07:002018-10-07T07:03:43.095-07:00hi , can you explain how you decide that MOAB is ...hi , can you explain how you decide that MOAB is cyclic quad.<br />thanks Anonymoushttps://www.blogger.com/profile/04192752888692172502noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-58108034092940263152018-07-17T12:43:10.163-07:002018-07-17T12:43:10.163-07:00https://photos.app.goo.gl/BGn1ZyJuoHQCZUsQ7
Connec...https://photos.app.goo.gl/BGn1ZyJuoHQCZUsQ7<br />Connect OA, OB, OC<br />Since AB and AC tangent to circle O => ∠ (BOA)= ∠ (COA)<br />We have ∠ (BFC)= ½. ∠ (BOC)= ∠ (BOA)<br />∠ (BFC)= ∠ (BMA)= ∠ (BOA) => MOAB is cyclic quad.<br />So ∠ (OMA)= ∠ (OBA) = 90 degrees => M is the midpoint of ED<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5109536389380223482018-07-16T23:52:50.532-07:002018-07-16T23:52:50.532-07:00< AMB = < BFC < ACB = < AOB since ABOC...< AMB = < BFC < ACB = < AOB since ABOC is concyclic<br /><br />Hence ABMO is concyclic and so < AMO = < ABO = 90<br /><br />Therefore M is the midpoint of DE<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.com