Geometry Problem

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

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Complete Problem 498

Level: High School, SAT Prep, College geometry

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## Wednesday, August 11, 2010

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

See also:

Complete Problem 498

Level: High School, SAT Prep, College geometry

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Geometry Problem

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

See also:

Complete Problem 498

Level: High School, SAT Prep, College geometry

Labels:
angle,
angle bisector,
double angle,
measurement,
triangle

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draw bisector of A, G point bisector meet BC

ReplyDelete1)▲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

since angle b is twice of angle c

ReplyDeletewe 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

INDSHAMAT,University of Kelaniya,Srilanka..

ReplyDeleteDraw 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.

ReplyDeleteAccording to Angle Bisector Theorem, 6/9=3/x,

therefore x=4.5.

Extend CA to E such that AB = AE = c

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

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

Sumith Peiris

Moratuwa

Sri Lanka