tag:blogger.com,1999:blog-6933544261975483399.post94438852970722640..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 835: Isosceles Triangle, Double Angle, Triple Angle, Auxiliary LinesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-21312070488801587622020-01-01T19:32:28.268-08:002020-01-01T19:32:28.268-08:00not clear at all. Your proposition is to
1) Drop ...not clear at all. Your proposition is to <br />1) Drop an altitude from C to the base, intercepting the base at F ( I think) that will bisect the vertex angle at C into 2 equal angles of 2beta each<br />2) The second construction I am not sure I understand. but it seems that you are proposing and angle bisector from A intersecting BC at E or intersecting DC At E.<br />the rest pointing to concurrency at G etc. I have no clue at all.<br />do you have a sketch of your construction?Anonymoushttps://www.blogger.com/profile/13687192779053288185noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12259446346041418072015-09-07T10:37:33.666-07:002015-09-07T10:37:33.666-07:00Let CX be the bisector of < ACB, X being on AD....Let CX be the bisector of < ACB, X being on AD. Now thro' congruence or symmetry < AXC = < BXC. But D is the in centre of Tr. ACX, so XD bisects < BXC. It thus follows that < BXD = 60. <br />So 2 alpha + 2 beta + 120 = 180 in Tr. ADC, implying that alpha + beta = 30.<br />So from Tr. BDC x = 150<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-83489741095154664502012-12-25T09:04:48.068-08:002012-12-25T09:04:48.068-08:00Fie P punctul de pe segmentul AD astfel incat m(: ...Fie P punctul de pe segmentul AD astfel incat m(: PC este mediatoarea segmetului AB si bisectoarea unghiului =>m(In ∆BPC <br />m(6α+6β=180 =>α+β=30 de unde in ∆BDC x=180-(α+β)=150<br /><br />Prof Radu Ion,Sc.Gim.Bozioru,Buzaunoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-75475401182796421492012-12-24T18:45:16.188-08:002012-12-24T18:45:16.188-08:00Minor typo correction.
Line 3 of my solution shoul...Minor typo correction.<br />Line 3 of my solution should read as "1. Draw altitude CF of triangle ABC and bisector AE of angle DAC "<br />Peter Tran<br /> Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-54490451121944137262012-12-23T13:35:56.532-08:002012-12-23T13:35:56.532-08:00http://img38.imageshack.us/img38/3597/problem835.p...http://img38.imageshack.us/img38/3597/problem835.png<br /><br />Make drawing per attached sketch<br />1. Draw altitude CF and bisector AF of angle DAC<br />Make angle ACE= beta<br />2. Due to symmetric properties , CDE and AEDB are isosceles<br />And BE, AD and CF are concurrent at point G<br />3. In triangle AGC , GE is an angle bisector => angle BGF=angle AGF=angle AGE=angle CGE=60<br />In triangle ABC we have 60+ 4alpha+4.beta=180 => alpha+ beta=30<br />In triangle BDC x=180-alpha-beta= 150<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-24079541743768594842012-12-23T12:44:22.596-08:002012-12-23T12:44:22.596-08:00And using elementary geometry?And using elementary geometry?Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67026801788334887692012-12-23T11:32:57.132-08:002012-12-23T11:32:57.132-08:00Result: 150
Solution: Let BC=AC=1 a:alfa, b:beta
...Result: 150<br /><br />Solution: Let BC=AC=1 a:alfa, b:beta<br />From sinus theorem in BDC<br />DC=sina/sin(a+b)<br />Sinus theorem in ADC using DC we get<br />sina*sin(2a+3b)=sin2a*sin(a+b) since sin2a=2*sina*cosa we get<br />2*sin(a+b)*cosa=sin(2a+b)+sinb=sin(2a+3b)<br />sinb=sin(2a+3b)-sin(2a+b)=2*sinb*cos(2a+2b)=sinb<br />eliminating sinb(b can't be zero) we get<br />cos(2a+2b)=1/2 which means 2a+2b=60, a+b=30<br />So x=180-30=150 degrees.Anonymousnoreply@blogger.com