Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemClick the figure below to see the complete problem 498 about Triangle, Angle Bisector, Double Angle, Measure.
draw bisector of A, G point bisector meet BC1)▲ABG ~ ▲ABC AB/9 = BG/AB2)AG bisector AB/BG = (3+x)/(9-BG)3)BD bisector AB/3 = 9/x=> x = 4,5
since angle b is twice of angle cwe have a equation a^2=c(c+a)ac=b^2-c^2 since ad is angular bisectord/a-d=c/bd=ac/b+cd=b^2-c^2/b+cd=b-c
INDSHAMAT,University of Kelaniya,Srilanka.. Draw the angle bisector of AngleBAC which cuts BC at E.Then AE=EC(as angleEAC=angleECA)3/x=AB/9(BD is the Bisector)So x=27/AB (1)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=4.5
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. According to Angle Bisector Theorem, 6/9=3/x,therefore x=4.5.
Extend CA to E such that AB = AE = cThen EB = ED so c+3 = 9 so c = 6Hence x/3 = 9/6 and so x = 4.5Sumith PeirisMoratuwaSri Lanka