tag:blogger.com,1999:blog-6933544261975483399.post8905599439927801524..comments2024-09-10T22:02:29.582-07:00Comments on GoGeometry.com (Problem Solutions): Elearn Geometry Problem 138 Nagel's TheoremAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-66541728949474423952013-09-17T15:41:06.143-07:002013-09-17T15:41:06.143-07:00Form 1:
O and H are isogonal conjugate, so the si...Form 1:<br /><br />O and H are isogonal conjugate, so the sides of pedal triangle of H are orthogonal to the respective line AO, BO or CO and reciprocally.<br /><br />Form 2:<br /><br />We know that A, B and C are the excenters of tr DEF, so by the well known existence of Bevan's point, we get that perpendiculars through A, B and C to the respective side of tr DEF are concurrent.<br /><br />Form 3:<br /><br />tr DEF is orthologic to tr ABC by definition, so by the orthology theorem, tr ABC must be orthologic to tr DEF.Editorhttps://www.blogger.com/profile/18079120609888942700noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41368184701973033742012-06-21T07:21:33.466-07:002012-06-21T07:21:33.466-07:00About problem 138.
In the solution by Vijay (March...About problem 138.<br />In the solution by Vijay (March 20, 2010), I didn't understand how the fact that angleAFE=C leads to the conclusion that AO is perpendicular do EF.<br />Can someone help me?Nilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-45946874690828280202010-12-09T08:16:24.530-08:002010-12-09T08:16:24.530-08:00The acute angle between the tangent at B to the ci...The acute angle between the tangent at B to the circumcircle and the chord BA equals the angle in the alternate segment angle BCA which in turn equals angle DFB (ACDF being a cyclic quadrilateral).<br />Therefore FD is parallel to the tangent at B and hence perpendicular to the radius BO.Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41742448631112838382010-03-20T07:07:05.952-07:002010-03-20T07:07:05.952-07:00angleBAO=angleABO=90-C and angleAHE=angleAFE(H is ...angleBAO=angleABO=90-C and angleAHE=angleAFE(H is ortho centre and AFHE is cyclic) but angle AHE=90-(90-C)=C so angleAFE=C hence AO is perpendicular to EF simillarly remaining we can provevijayhttp://vijay.comnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-42747804659642018682008-09-06T01:08:00.000-07:002008-09-06T01:08:00.000-07:00as O is the circumcenter,∠OBC=90°-∠CAB=∠ABE.as FAC...as O is the circumcenter,<BR/>∠OBC=90°-∠CAB=∠ABE.<BR/>as FACD concyclic,<BR/>∠BDF=∠CAB.<BR/>thus BO⊥DF is obvious, so are the others.Anonymousnoreply@blogger.com