Wednesday, August 11, 2010

Problem 498: Triangle, Angle Bisector, Double Angle, Measure

Geometry Problem
Click the figure below to see the complete problem 498 about Triangle, Angle Bisector, Double Angle, Measure.

Problem 498. Triangle, Angle Bisector, Double Angle, Measure.


See also:
Complete Problem 498

Level: High School, SAT Prep, College geometry

5 comments:

  1. draw bisector of A, G point bisector meet BC
    1)▲ABG ~ ▲ABC AB/9 = BG/AB
    2)AG bisector AB/BG = (3+x)/(9-BG)
    3)BD bisector AB/3 = 9/x

    => x = 4,5

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  2. since angle b is twice of angle c
    we have a equation a^2=c(c+a)
    ac=b^2-c^2
    since ad is angular bisector
    d/a-d=c/b
    d=ac/b+c
    d=b^2-c^2/b+c
    d=b-c

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  3. 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

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  4. 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.

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  5. Extend CA to E such that AB = AE = c

    Then EB = ED so c+3 = 9 so c = 6

    Hence x/3 = 9/6 and so x = 4.5

    Sumith Peiris
    Moratuwa
    Sri Lanka

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