tag:blogger.com,1999:blog-6933544261975483399.post233720196256502909..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 184Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-6933544261975483399.post-43297222169755054662020-12-09T18:43:44.974-08:002020-12-09T18:43:44.974-08:00That BXZY is concyclic (with BY diameter) is clear...That BXZY is concyclic (with BY diameter) is clear because Tr.s ABX & ADX are congruent SSS, <BXZ = 90 = <BZD (by construction)<br /><br />Now BDZ is a 30-60-90 triangle so DXZ is equilateral<br /><br />Now Y lies on the perpendicular bisector of BD and hence BY = DY. But < BDY = 60 so Tr. BDY is equilateral<br /><br />Trust this clarifiesSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-9208830258072730192020-12-09T05:43:08.751-08:002020-12-09T05:43:08.751-08:00Sir Mr.Sumith, Please may I ask; kindly explain ho...Sir Mr.Sumith, Please may I ask; kindly explain how are <br />"BXZY is concyclic and hence Tr.s DXZ and BDY are both equilateral".<br />Thanking you in advance.Maung Theinhttps://www.blogger.com/profile/14206850669851100359noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74511898162940648982020-05-02T20:32:28.171-07:002020-05-02T20:32:28.171-07:00It is special case of problem (t, 3t, 30) where t ...It is special case of problem (t, 3t, 30) where t = 16.<br /><br />It is the same problem with following cevian problem (Circumcircle BCD meets AC at P)<br />https://output.jsbin.com/fofecum#48,16,14,44 <br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-9752775354907765382019-04-09T01:24:57.888-07:002019-04-09T01:24:57.888-07:00Let me rephrase "Let DZY be perpendicular to ...Let me rephrase "Let DZY be perpendicular to BC, Z on BC and Y on AX extended." <br /><br />"Drop a perpendicular from D to BC to meet BC at Z. AX and DZ extended meet at Y"<br /><br />Trust this clarifiesSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-51026589827663916142019-04-02T10:14:43.332-07:002019-04-02T10:14:43.332-07:00Sumith: I do not understand "Let DZY be perpe...Sumith: I do not understand "Let DZY be perpendicular to BC, Z on BC and Y on AX extended." Would you kindly elaborate or provide a diagram please.Ercan Cemhttps://www.blogger.com/profile/17715448112482768479noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13603842937444403922019-02-15T00:20:17.538-08:002019-02-15T00:20:17.538-08:00Let X be midpoint of BD. Let DZY be perpendicular ...Let X be midpoint of BD. Let DZY be perpendicular to BC, Z on BC and Y on AX extended. <br /><br />BXZY is concyclic and hence Tr.s DXZ and BDY are both equilateral. <br />So BX = DX = XZ = DZ = ZY and hence CD = CY. <br /><br />Now mark P on AY such that CP = CY (= CD). So < CYP = < CPY = < CDQ where Q is on AD extended. <br /><br />It follows that ADCY is concyclic and x = < AYD = 30<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-32962433094539797092016-06-10T07:27:23.970-07:002016-06-10T07:27:23.970-07:00Problem 184
Forming an equilateral triangle ADE (E...Problem 184<br />Forming an equilateral triangle ADE (E is above the AC),then AB=AE=AD=ED .But <EAB=4,<br /><EBA=(180-4)/2=88=58+30, so<EBD=30. Then B, E,C are collinear.But <EDB=2,<DEC=32, so <br /><EAC=44=<ECA ,so EC=AE.Therefore E is the center of the circle triangle ADC so <br /><ACD=<AED/2=60/2=30.<br />APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-87193178008679578062009-04-09T22:10:00.000-07:002009-04-09T22:10:00.000-07:00Angle ABD=58 which makes angle ABC = 88 and angle ...Angle ABD=58 which makes angle ABC = 88 and angle ACB =44. AC/AD=sin(x+16)/sin(x) =AC/AB=sin(88)/sin(44) which leads us to a simple equation like: sin(x+16)/sin(x) = 2cos(44) which gives x=30 deg since sin(46)=cos(44). Thus, x=30 deg.<BR/>I still would like to know how this done by plane geometry alone.<BR/>Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-82238980280541181272008-10-31T18:02:00.000-07:002008-10-31T18:02:00.000-07:00Yes, I have a geometric solution.Thank you for you...Yes, I have a geometric solution.<BR>Thank you for your question.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-90150338110067092002008-10-31T17:34:00.000-07:002008-10-31T17:34:00.000-07:00x=30. I have an algebraic proof. Antonio, do you h...x=30. I have an algebraic proof. Antonio, do you have a geometric one?<BR/><BR/>Orange Skidorangeskidhttps://www.blogger.com/profile/10938522388786613059noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-16152333451740368822008-10-12T14:06:00.000-07:002008-10-12T14:06:00.000-07:00Evrenin, the answer x = 32 is not correct, try aga...Evrenin, the answer x = 32 is not correct, try again. Good luck.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-19059587817929971072008-10-12T12:21:00.000-07:002008-10-12T12:21:00.000-07:00x=32x=32Evren Topcuhttps://www.blogger.com/profile/12531378528239540978noreply@blogger.com