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