tag:blogger.com,1999:blog-6933544261975483399.post261557953897632579..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 964: Right Triangle, Cevians, Angles, 10, 20, 30 DegreesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-31228515813045165672022-11-06T19:08:28.012-08:002022-11-06T19:08:28.012-08:00Con Encentró sale inmediatoCon Encentró sale inmediatoAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-62319435759185001722017-01-02T22:41:27.557-08:002017-01-02T22:41:27.557-08:00Alternatively one could use trigonometry. Angle DE...Alternatively one could use trigonometry. Angle DEC = 10 deg, so CD=DE. Let CD=1 (unit length). AD = DE*cos(20 deg), BA=AC*tan(40 deg). DE=1, AC=1+AD=1+cos(20 deg)<br />Tan(angle ABD) = AD/BA=cos(20deg)/(1+cos(20 deg))/tan(40 deg) = tan(30) <br />Angle ABD = 80 - X = 30. So X = 50 deg.Anonymoushttps://www.blogger.com/profile/16146754496953151724noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-56494069033334315042015-07-06T09:59:56.554-07:002015-07-06T09:59:56.554-07:00Excellent solution FUKUCHANExcellent solution FUKUCHANSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5156947004833506002014-01-27T16:05:42.328-08:002014-01-27T16:05:42.328-08:00See http://fukuchandayo.web.fc2.com/GG_964.html
A...See http://fukuchandayo.web.fc2.com/GG_964.html<br /><br />Angle DEC=10 deg. ---> triangle DCE is isoceles ---> CD=DE.<br />Put point G on BC, so that DC=DG ---> angle DGC=DCG=40 deg.<br />---> angle CDG=100 deg. ---> angle GDE=60 deg.<br />Triangle DGE is equilateral <--- DE(=CD)=DG.<br />Angle GEB=180-70-60=50 deg.=angle GBE.<br />---> GE=GB, and angle BGE is 80 deg.<br /><br />Points B, E, D are on circle G.<br />---> angle BDE=angle BGE/2=40 deg.<br /><br />Angle BFE(=x)=40+10=50 deg.FUKUCHANnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-42840961186167221822014-01-27T09:29:34.732-08:002014-01-27T09:29:34.732-08:00Consider G pe BC astfel incatDG= ED=DC,rezulta E...Consider G pe BC astfel incatDG= ED=DC,rezulta EDG echilateral,EGB,GDC,BGD isoscele=>BG=EC=GD=ED=DC m<BDG=20°,m<BDE=40°.x=50°.ion radunoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-53821976237129602372014-01-26T22:02:36.088-08:002014-01-26T22:02:36.088-08:00I do not have a full solution but a suggestion for...I do not have a full solution but a suggestion for others. Consider G the midpoint of BD. G is the circumcentre of Tr. ABD or GA=GD. For reasons that are still unclear to me, GAD is an equilateral triangle, In other words, AD=GA=GD or AD=BD/2 which in turn suggests that BAD is a 30°-60°-90° triangle. /_EDF is therefore=40° and thus x=(40+10)=50°.<br />All that remains now is to prove that Tr. GAD is equilateral. Still beyond me though -- A lovely problem, Antonio!Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com