Wednesday, February 18, 2009

Problem 11: Right triangle, Cevian, Angles

Proposed Problem

See complete Problem 11 at:
http://gogeometry.com/problem/problem011.htm

Level: High School, SAT Prep, College geometry

1. Triangles DAB & ACB are similar. Hence, BD/AB = AB/BC or AB^2 = BD*BC = 3BD^2. In triangle DAB, we've AD^2 = AB^2+BD^2 = 3BD^2+BD^2 = 4BD^2. Thus, BD = AD/2. In other words, DAB is 30-60-90 with angle DAB = x = 30 deg.
Ajit: ajitathle@gmail.com

2. Solucion:
http://img14.imageshack.us/img14/6471/rsolmarzo6.jpg

3. http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf

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5. Let AB=c and DC=2a=2BD
From triangle ABD and ABC:tanx=a/c=c/3a
so 3a^2=c^2 or a/c=(sqrt3)/3 So tanx=(sqrt3)/3
and x=30 deg

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7. My solution

http://lectiimatematice.blogspot.com/2012/02/var-docstocdocid112425849var.html

9. in triangle ABC, tanx= AB/BC....In triangle ABD, tanx =BD/AB.....Multipying both we get x=30

11. Extend CB to E such that BD = BE.

Then AB is the perpendicular bisector of DE and < EAB = < BAD = x = < ACE.
Hence < EAC must be a right angle and so AD=DE=DC
Therefore < DAC = x, hence 3x = 9and x =30

Sumith Peiris
Moratuwa
Sri Lanka

12. Triangle ABD is similar to triangle ABC

Therefore, BD/AB = AB/3BD
AB^2 = 3BD^2
AB = (sqr3)BD

tanx = BD/(sqr3)BD
x = tan^-1(1/sqr3)
Therefore x = 30

13. h/BD=3BD/h, h^2= 3BD^2, BD/h= 1/sqrt(3)-> 30°

14. Draw the circle circumscribed to ∆ABC, centered on H, midpoint of AC.
Extend AD to cut circle at E.
Then, subtending same arc BE: <BAE=BCE=x, thus <ACE=2x
Therefore, in right triangle AEC : <EAC+<ACE=3x=90°, finally x=30°

15. The solution for this is mostly arithmetic.

Firstly, notice that triangles ABD and ABC are similar since they have the same angles.

This means that they have identical side-side ratios, no matter their lengths.

As an example, visualise a isosceles right-angled triangle. No matter the size of the right triangle, the ratio of any leg to the hypotenuse is always 1:√2

Therefore,
BD/AB = AB/BC
BD/AB = AB/(BD + 2BD)
1/AB = AB/3
AB/AB^2 = AB/3
AB^2 = 3
AB = √3 BD

These are properties are those of half an equilateral triangle, since the altitude of equilateral triangle is always √3 of half its length.

Therefore, x = 60°/2 = 30°

16. Let BD=a, DC=2a
Consider triangle ABD
tanx=a/BA
BA=a/tanx
Consider triangle ABC
tanx=AB/BC
=a/3atanx
tan^2(x)=1/3
tanx=1/sqrt(3)
x=30
tanx

17. Extend DB to E such that BE=BD => AED is isosceles
m(EAB)=x=> m(EAC)=90 and D is the circumcenter of Right Tr. EAC => m(DAC)=m(DCA) => 90-2X=X. Hence X =30