## 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. Complete Problem 498

Level: High School, SAT Prep, College geometry

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

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

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

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.

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