Proposed Problem
See complete Problem 11 at:
http://gogeometry.com/problem/problem011.htm
Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Wednesday, February 18, 2009
Problem 11: Right triangle, Cevian, Angles
Subscribe to:
Post Comments (Atom)
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.
ReplyDeleteAjit: ajitathle@gmail.com
Solucion:
ReplyDeletehttp://img14.imageshack.us/img14/6471/rsolmarzo6.jpg
http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteLet AB=c and DC=2a=2BD
ReplyDeleteFrom 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
This comment has been removed by the author.
ReplyDeleteMy solution
ReplyDeletehttp://lectiimatematice.blogspot.com/2012/02/var-docstocdocid112425849var.html
The solution is uploaded to the following link:
ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJZmw0MVhmWHFFLVE
in triangle ABC, tanx= AB/BC....In triangle ABD, tanx =BD/AB.....Multipying both we get x=30
ReplyDeletehttps://www.youtube.com/watch?v=ofDdrHRObOI
ReplyDeleteExtend CB to E such that BD = BE.
ReplyDeleteThen 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
Triangle ABD is similar to triangle ABC
ReplyDeleteTherefore, BD/AB = AB/3BD
AB^2 = 3BD^2
AB = (sqr3)BD
tanx = BD/(sqr3)BD
x = tan^-1(1/sqr3)
Therefore x = 30
See original solution on this video
ReplyDeleteSee original solution on this video
ReplyDeleteh/BD=3BD/h, h^2= 3BD^2, BD/h= 1/sqrt(3)-> 30°
ReplyDeleteDraw the circle circumscribed to ∆ABC, centered on H, midpoint of AC.
ReplyDeleteExtend 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°
The solution for this is mostly arithmetic.
ReplyDeleteFirstly, 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°
Let BD=a, DC=2a
ReplyDeleteConsider 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
Extend DB to E such that BE=BD => AED is isosceles
ReplyDeletem(EAB)=x=> m(EAC)=90 and D is the circumcenter of Right Tr. EAC => m(DAC)=m(DCA) => 90-2X=X. Hence X =30