## Wednesday, February 18, 2009

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

Proposed Problem

See complete Problem 15 at:
http://gogeometry.com/problem/problem015.htm

Level: High School, SAT Prep, College geometry

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

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

3. bae deok rak(bdr@korea.com)October 4, 2010 at 8:05 PM

Let 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.

5. 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
So 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