tag:blogger.com,1999:blog-6933544261975483399.post1950720879799783499..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1240: Triangle, Circle, Diameter, Circumcircle, Circumcenter, Perpendicular. Mobile AppsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-80135091390547687212019-12-23T06:10:49.038-08:002019-12-23T06:10:49.038-08:00Let (Q) be the circle with diameter AC, and Bt the...Let (Q) be the circle with diameter AC, and Bt the tangent to circle (O) on B.<br />Circles (Q) and (O) intersect at D and E.<br />Line (ADB) cuts (Q) on A and (O) on B.<br />Line (ECB) cuts (Q) on C and (O) on B. Applying Reim’s theorem:<br />AC//Bt and since Bt┴BO BO┴AC.<br />Gewürtztraminer68https://www.blogger.com/profile/00329980114313015297noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-15133644857930371552016-08-01T09:53:10.684-07:002016-08-01T09:53:10.684-07:00let angle CAB = A, ABC = B and ACB = C
=> angl...let angle CAB = A, ABC = B and ACB = C<br /><br />=> angle DEC = DEB = DAC = A => triangles ABC & EBD are similar<br /><br />Extend BO to meet the circle at F and join FE<br />=> <FED = 90+A <br /><br />Since FEDB is cyclic quadrilateral <DBF = <DBO = <ABO = 90-A<br /><br />So if a triangle is formed by extending AC to intersect BO at a point P - it is right angle at P as <PAB = A & PBA = 90-AAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5296000388363417522016-07-30T22:34:00.999-07:002016-07-30T22:34:00.999-07:00https://goo.gl/photos/LyXa4bZgSQcpQGuR9
Draw diam...https://goo.gl/photos/LyXa4bZgSQcpQGuR9<br /><br />Draw diameter BF of circle O , lines AE and CD<br />Since AD⊥DC= > DC will cut circle O at F<br />Since AE⊥CE => AE will cut circle O at F<br />In triangle ABF , C is the orthocenter so AC⊥BO<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-46669546782430515102016-07-30T21:24:23.220-07:002016-07-30T21:24:23.220-07:00Correction to last line.......
If BO meets AC at ...Correction to last line.......<br /><br />If BO meets AC at X, it follows that DBXC is concyclic and < CDA being 90, < BXC is also = 90Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-91112508266585500162016-07-30T20:01:10.527-07:002016-07-30T20:01:10.527-07:00If < DAC = A then < DEB = A and < DOB = 2...If < DAC = A then < DEB = A and < DOB = 2A<br /><br />Now DOB being isoceles < DBO = 90-A<br />But < DCA = 90-A<br /><br />It follows that DBOC is concyclic and < CDA being 90, < BOC is also = 90<br /><br />Sumith Peiris<br />Moratuwa<br />Sri Lanka Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.com