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

Online Geometry theorems, problems, solutions, and related topics.

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