## 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