Proposed Problem

See complete Problem 15 at:

http://gogeometry.com/problem/problem015.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Wednesday, February 18, 2009

### Problem 15: Triangle, Cevian, Angles, Congruence

Labels:
angle,
cevian,
congruence,
triangle

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Extend BA to point E such that AE=AD. Now,

ReplyDeleteAB+AD = AB+AE = BE. But BC=AB+AD. Hence BE=BC. In triangles EBD & CDB, AD is common, the included angles ABD & DBC are each = 30 and BE=BC which makes the triangles congruent. Thus angle AED = x and angle ADE also =x since AD=AE by construction while angle BAD=x+x=2x and sngle ADB=x+30. Triangle ABD now gives, 30+2x+30+x=180 so x=40 deg.

Ajit: ajitathle@gmail.com

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

ReplyDeleteLet E be a point on BC with BE=BA. Since BA=BE and ang(ABE)=60 deg,triangle BAE is an Equilateral triangle and DA=DE=EC. Hence we see that ang(DCE)=x=ang(CDE) and ang(DEA)=x/2 and hence 2x=x/2+60, that is , x=40 degree.

ReplyDeleteThe solution is uploaded to the following link:

ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJMk52SXpZdlZFRFU

Extend BA to F such that BF = BC. Hence Tr. FBC is equilateral and both Tr.s FAD & FDC are isoceles since BD is the perpendicular bisector of FC

ReplyDeleteSo from Tr. FDC 60-x= 60-A/2 and so x= A/2, which can only happen when A= 80 and x=40 since B= 60

Sumith Peiris

Moratuwa

Sri Lanka

https://www.youtube.com/watch?v=Rg1AM58ShnE

ReplyDelete