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

Labels:
angle,
cevian,
congruence,
right triangle

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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