tag:blogger.com,1999:blog-6933544261975483399.post1746064253098379394..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 498: Triangle, Angle Bisector, Double Angle, MeasureAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-75205804377690955882023-12-21T22:45:22.643-08:002023-12-21T22:45:22.643-08:00As (<BCA), which implies that AD should be grea...As (<BCA), which implies that AD should be greater than CD which is not reasonable, so such triangle actually does not exist.Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-16842927231676278832016-06-21T15:39:43.924-07:002016-06-21T15:39:43.924-07:00Extend CA to E such that AB = AE = c
Then EB = ED...Extend CA to E such that AB = AE = c<br /><br />Then EB = ED so c+3 = 9 so c = 6<br /><br />Hence x/3 = 9/6 and so x = 4.5<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-90832240712001421932010-10-23T10:39:05.327-07:002010-10-23T10:39:05.327-07:00Just one segment drawn (DE, E on BC so Angle CDE=A...Just one segment drawn (DE, E on BC so Angle CDE=Angle DCE=a, making BED=2a, so Triangles ABD and EBD) reveals AB=6.<br /> According to Angle Bisector Theorem, 6/9=3/x,<br />therefore x=4.5.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-20594094280708087972010-09-07T11:21:12.446-07:002010-09-07T11:21:12.446-07:00INDSHAMAT,University of Kelaniya,Srilanka..
Draw...INDSHAMAT,University of Kelaniya,Srilanka..<br /> Draw the angle bisector of AngleBAC which cuts BC at E.Then AE=EC(as angleEAC=angleECA)<br />3/x=AB/9(BD is the Bisector)So x=27/AB (1)<br />angleAEB=2@ and because angleABC is common for both ABE & ABC triangles,ABC & ABE triangles are similar.Hence AE/(x+3)=CE/(x+3)=AB/9=BE/AB(2).So BE=AB2/9,hence CE=9-BE=(81-AB2)/9(3).From(2)we have CE/(x+3)=AB/9.So applying values of(3)&(1)to CE&DC we have 81-AB2=3(9+AB).So (9-AB)(9+AB)=3(9+AB).So AB=6.Hence from(1)x=27/AB=27/6=9/2,So x=9/2=4.5INDSHAhttps://www.blogger.com/profile/13177164616312631411noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63057042145807280092010-09-05T06:13:22.068-07:002010-09-05T06:13:22.068-07:00since angle b is twice of angle c
we have a equati...since angle b is twice of angle c<br />we have a equation a^2=c(c+a)<br />ac=b^2-c^2 <br />since ad is angular bisector<br />d/a-d=c/b<br />d=ac/b+c<br />d=b^2-c^2/b+c<br />d=b-cvishnu thejanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-61011547818770494362010-08-11T08:46:27.531-07:002010-08-11T08:46:27.531-07:00draw bisector of A, G point bisector meet BC
1)▲AB...draw bisector of A, G point bisector meet BC<br />1)▲ABG ~ ▲ABC AB/9 = BG/AB<br />2)AG bisector AB/BG = (3+x)/(9-BG)<br />3)BD bisector AB/3 = 9/x<br /><br />=> x = 4,5c .t . e. onoreply@blogger.com