tag:blogger.com,1999:blog-6933544261975483399.post8117577276218353389..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 210: Triangle, Angles, Auxiliary linesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger22125tag:blogger.com,1999:blog-6933544261975483399.post-71909527297509949352023-03-04T22:50:25.440-08:002023-03-04T22:50:25.440-08:00[For easy typing, I use b instead of alpha]
Consid...[For easy typing, I use b instead of alpha]<br />Consider Triangle BDC<br /><BDC=60-b<br />sin(120-2b)/DC=sin(60-b)/a<br />DC/a=2cos(60-b)---------(1)<br /><br />Consider Triangle DEC<br /><DEC=180-x-b<br />sin(180-x-b)/DC=sinx/a<br />DC/a=sin(x+b)/sinx---------(2)<br /><br />By equating (1) and (2)<br />2cos(60-b)sinx=sin(x+b)<br />sin(x+60-b)+sin(x+b-60)=sin(x+b)<br />sin(x+60-b)=sin(x+b)-sin(x+b-60)<br />sin(x+60-b)=2cos(x+b-30)sin30<br />sin(x+60-b)=cos(x+b-30)<br />sin(x+60-b)=sin(120-x-b)<br />x+60-b=120-x-b<br />x=30Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-45350806018206262912020-12-30T06:24:07.240-08:002020-12-30T06:24:07.240-08:00With a new link that works to access the diagram....With a <b><a href="https://drive.google.com/file/d/1R6_9Tku13EnJzuUt1JIC27MSrp0JqAQI/view?usp=sharing" rel="nofollow"> new link</a></b> that works to access the diagram.Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-3198248247730156942020-12-30T06:14:18.635-08:002020-12-30T06:14:18.635-08:00With a link that works to see the diagram.With a <b><a href="https://drive.google.com/file/d/17J1nTi-5Y3YBSMDNrmGaMj_q3nGlq3qe/view?usp=sharing" rel="nofollow"> link</a></b> that works to see the diagram.Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-73247707290130485782020-11-29T01:40:54.928-08:002020-11-29T01:40:54.928-08:00In the above solution, if α < 15°, ∠ABC is obtu...In the above solution, if α < 15°, ∠ABC is obtuse, F lies on AB extended and segment CB lies within ∠ECF.<br />In such case, ∠BCF = 60°- 4.α ⇒ ∠FCE = 60° and the remaining of the proof stays the same.<br /><b>x = 30°</b><br />Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-6143262043580010512020-11-28T08:02:58.011-08:002020-11-28T08:02:58.011-08:00Another solution : see graph here.
Let F on AB s...Another solution : see graph <a href="https://photos.google.com/photo/AF1QipMLH2HopuzE2qYhuNYGX6bvSt9_4cHW6EfdyQn3" rel="nofollow"> here</a>.<br /><br />Let F on AB such that CF = CB = a.<br /><br />∆BCF isoscele in C ⇒ ∠BCF = 4.α– 60° ⇒ ∠FCE = 60° ⇒ ∆FCE is equilateral.<br /><br />In ∆CDF, ∠CDF = 60°- α = ∠CDF ⇒ ∆CDF isoscele in F ⇒ FD = FC = FE = a.<br /><br />So F is the center of the circumcircle to ∆CDE and ∠CFE/2 = ∠CDE = x = 30°Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-52672458916916653302020-11-28T06:49:22.605-08:002020-11-28T06:49:22.605-08:00See graph here.
BCE isoscele in C and ∠C = 4.�� ⇒...See graph <a href="https://photos.google.com/photo/AF1QipOldIKnmmTg0AtPVlls-zuDqE2Ef8ML8kU1B5jo" rel="nofollow"> here</a>.<br />BCE isoscele in C and ∠C = 4.�� ⇒ ∠DBE=30°.<br />The perpendicular to AB from E intersects AB in F and the angle bisector of ∠BCE in G<br />The perpendicular to CD from E intersects CD in H and CG in I.<br />Now ∠EHD = ∠EFD = 90° ⇒ F and H are on the circle of diameter ED ⇒ ∠EDH = ∠EFH.<br />F middle of GE (easily seen since ∠BEF=60°) and H middle of EI ⇒ FH // GI and ∠EFH = ∠EGI.<br />So ∠EDC = ∠EDH = ∠EFH = ∠EGI = 30° => x = 30°.Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-49451510942923138712019-02-15T00:01:53.132-08:002019-02-15T00:01:53.132-08:00Let X be midpoint of BE. Let the perpendicular to ...Let X be midpoint of BE. Let the perpendicular to AB drawn from E meet CX extended at Y and AB at Z.<br /><br />Since BXZY is concyclic and < ZBX = 30 (since < CBE = 90- 2@), < XYZ = < XBY = 30 and Tr.s EXZ and BEY are both equilateral. Also BX = EX = XZ = EZ = ZY<br /><br />So DE = DY. Now since EY is perpendicular to DB, EY cannot be perpendicular to CD so mark P on CY such that EP is perpendicular to CD.<br /><br />This yields DE = DP = DY and < DYP = < DPY = DEA, hence CEDY is concyclic<br /><br />In turn x = < CYE = 30<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-56410056450773822242016-08-09T03:52:11.236-07:002016-08-09T03:52:11.236-07:00Problem 210
Let point P in the extension of AB to...Problem 210<br />Let point P in the extension of AB to B such that CB=CP=CE (if α<15).Τhen 15.<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-81866600117798858082013-02-01T02:58:23.351-08:002013-02-01T02:58:23.351-08:00Thanks Antonio.
Got it.
Thanks Antonio.<br />Got it.<br /><br />Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-55581418197023351692013-01-31T07:22:54.320-08:002013-01-31T07:22:54.320-08:00Dear Ajit, Thanks for your comment. In general cas...Dear Ajit, Thanks for your comment. In general case x = 30 degrees. Hint for a geometric prove: draw CF = CB (F on AB extended)Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-30121742208411437692013-01-28T19:52:23.009-08:002013-01-28T19:52:23.009-08:00Antonio,
If alpha=15 deg. then clearly x=30 deg. ...Antonio, <br />If alpha=15 deg. then clearly x=30 deg. and this can be done w/o trigonometry.<br />However, in the general case, is x =2*(alpha)? And how may we prove it?<br />Please enlighten us.<br />A²Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18136020160306706232010-05-03T12:36:01.306-07:002010-05-03T12:36:01.306-07:00fie N pe BD astfel iancat CN=a.Se obtine triunghiu...fie N pe BD astfel iancat CN=a.Se obtine triunghiul NEC echilateral, triunghiul DNC isoscel, triunghiul DNE isoscel si se calculeaza unghiul EDC=30.nicknoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-60625450836741256862009-10-31T07:18:51.550-07:002009-10-31T07:18:51.550-07:00http://geometri-problemleri.blogspot.com/2009/10/p...http://geometri-problemleri.blogspot.com/2009/10/problem-39-ve-cozumu.htmlAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-54677984420341907972009-03-26T12:52:00.000-07:002009-03-26T12:52:00.000-07:00From where did you obtain BCEN is an inscribed qua...From where did you obtain BCEN is an inscribed quadrilateral ? , and why did you say "let N be a point on BD" , and afterwards you defined it with the angle MEN = 30º ? ok , let's say no problem there but , I repeat from W(here)TF did you obtain BCEN is an inscribed quadrilateral , it's an particular case , i think the real answer is : 2*ALPHA .<BR/>Check other cases and it's not 30 degrees .AurelMVAhttps://www.blogger.com/profile/01140777294986346786noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-7448168378546508892009-03-25T12:21:00.001-07:002009-03-25T12:21:00.001-07:00why #BCEN is an inscribed quadrilateral?why #BCEN is an inscribed quadrilateral?Unknownhttps://www.blogger.com/profile/17597446174103329489noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-73165496173910051562009-03-22T11:09:00.000-07:002009-03-22T11:09:00.000-07:00Firstly, I trace the line BE. Because the triangle...Firstly, I trace the line BE. Because the triangle BCE is isosceles(CB = CE), I can deduce that angle DBE = 30º. Now, Let M be a point on the midle point of segment BE(MB = ME)and let N be a point on BD. Then, CM is the height of tringle BCE. Now, because I can deduce that the angle MEN = 30º = DBE, I see that the triangle BNE is isoceles(NB = NE). Now I can deduce that the points B, C, E and N delimit the quadrilateral BCEN which is an inscribed quadrilateral inside an circle. As the points C, M and N are on the same line and the CM is the height of the isosceles triangle BCE, the line CN is the diameter of this circle. Then, the angle NBC = 90º and ABC is a right(or pytagorean) triangle. So, 120 - 2alpha = 90º and then alpha = 15º. Then, I can easily deduce that DBC is an right triangle(BC = BD and angle BDC = angle BCD =45º)and that BCE is an equilateral triangle. Now, I can see that triangle BDE is isosceles(BD = BE). As the angle DBE = 30º, then the angle EDB = 75º. Finally, the measure of the angle CDE = angle EDB - angle BDC = 75º - 45º = 30ºAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27037474786801957822008-12-22T20:01:00.000-08:002008-12-22T20:01:00.000-08:00Dear Antonio,In fact I can`t to say at first that ...Dear Antonio,<BR/><BR/>In fact I can`t to say at first that angle DBC + angle DEC = 180º. So, my answear at first should be wrong. I`l search better.Unknownhttps://www.blogger.com/profile/08069391683750435104noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-3262639902827791192008-12-22T18:48:00.000-08:002008-12-22T18:48:00.000-08:00Dear Jandir,Thank you for your comments.My answer:...Dear Jandir,<BR/>Thank you for your comments.<BR/>My answer:<BR/>1. Why angle DBC + angle DEC = 180º?<BR/>2. You don't need more data to solve this problem.<BR/>3. In fact, you can deduce (it's not a data) that the angle between BE and BD measured 30 degrees.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63354908266376549622008-12-22T18:45:00.000-08:002008-12-22T18:45:00.000-08:00Dear bolchoi,Thank you for your comments.My answer...Dear bolchoi,<BR/>Thank you for your comments.<BR/>My answer:<BR/>1. Why angle DBC + angle DEC = 180º?<BR/>2. You don't need more data to solve this problem.<BR/>3. In fact, you can deduce (it's not a data) that the angle between BE and BD measured 30 degrees.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33817921334015942232008-12-22T18:38:00.000-08:002008-12-22T18:38:00.000-08:00Okay Antonio. Thanks, I could to sove this problem...Okay Antonio. Thanks, I could to sove this problem. So, All data are correct ! <BR/>angle DBC + angle DEC = 180º. So, quadrilateral BCDE is inscribed. Now, the angle between the auxiliary line BE and the segment BD is 30º. Then, the angle DBE = DCE = a = 30°. On the other hand, angle EDC(=X)is equal to EBC = 90 -2a. So, X= 90 - 2(30)= 30º.<BR/><BR/>Bolchoi(from Brazil)Unknownhttps://www.blogger.com/profile/08069391683750435104noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-7930742208571995002008-12-22T17:08:00.000-08:002008-12-22T17:08:00.000-08:00The angle between the auxiliary line BE and the se...The angle between the auxiliary line BE and the segment BD is 30º. Does the data is enought o solve this problem ??Unknownhttps://www.blogger.com/profile/11270649438404611442noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-75084838229537396432008-12-12T12:50:00.000-08:002008-12-12T12:50:00.000-08:00the solve 30[img]http://www.al3ez.net/upload/d/ibr...the solve 30<BR/>[img]http://www.al3ez.net/upload/d/ibrahim_ibrahim_page4.jpg[/img]ibrahim elsagerhttps://www.blogger.com/profile/08770404937202824228noreply@blogger.com