tag:blogger.com,1999:blog-6933544261975483399.post6028467590586330966..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 198: Triangle, Quadrilateral, AngleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-88541590468534667222015-07-28T03:15:17.187-07:002015-07-28T03:15:17.187-07:00Draw altitudes BP and BQ to DC and AC respectively...Draw altitudes BP and BQ to DC and AC respectively which are easily seen to be equal since BC bisects < ACP. <br />As a result right Tr.s ABQ and DBP are congruent from which it follows that < BAC = < BDC. Hence ABCD is cyclic and x = 50.Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5771339634961100592012-08-08T17:41:18.718-07:002012-08-08T17:41:18.718-07:00To Peter, about problem 198.
Thank you for your he...To Peter, about problem 198.<br />Thank you for your help. Now I understand the solution by Jankonyex.<br />I wish to propose other solution, I hope it's right.<br /><br />Let O be the circumcenter of triangle ACD. The straight line BO, which is the line bisector of AD, cuts that circle at P, that is midpoint of the arc AD. So CP is angle bisector of ACD. Then we have ang(BCP) = 65º + 25º = 90º.<br />The circle O is also circumscribed to CDP, so the perpendicular OQ to CP is line bisector of CP and CQ = QP.<br />But OQ is parallel to BC so O is midpoint of BP. This means that B belongs to the circle O<br />and ABCD is cyclic. Thus x = ang(ABD) = ang(ACD) = 50º.Nilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13483065030629588612012-08-07T21:03:57.681-07:002012-08-07T21:03:57.681-07:00Nilton
See sketch below for detail.
http://img69...Nilton<br /><br />See sketch below for detail. <br />http://img692.imageshack.us/img692/6971/problem198c.png<br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88219559414877300412012-08-06T18:22:31.993-07:002012-08-06T18:22:31.993-07:00About Jankonyex's solution for problem 198.
Co...About Jankonyex's solution for problem 198.<br />Could anybody tell me why is ACD' a straight line?<br />I think that this explanation is missing.<br />Thanks.Nilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12190081443846174012008-10-29T02:07:00.000-07:002008-10-29T02:07:00.000-07:00better solution:maps D to D' by reflecting about B...better solution:<BR/>maps D to D' by reflecting about BC<BR/>ACD' is a straight line, BAD' is an isos. triangle<BR/>mCAB=mBD'C=mCDB, and hence x=50°Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70663258182775129172008-10-28T03:41:00.000-07:002008-10-28T03:41:00.000-07:00maps D to D' by reflecting about BCB is the circum...maps D to D' by reflecting about BC<BR/>B is the circumcenter of ADD', ACD' is a straight line and CDD' is an isos. triangle<BR/>x=2mCD'D=50°Anonymousnoreply@blogger.com