See complete Problem 227 at:

gogeometry.com/problem/p227_triangle_centroid_perpendicular.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Tuesday, January 20, 2009

### Elearn Geometry Problem 227: Triangle, Centroid, Exterior line

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Let DF be px+qy+k=0 and let A:(0,0),B:(a,b) & C:(c,0). Therefore,d=k/(p^2+q^2)^(1/2),e=(ap+bq+k)/(p^2+q^2)^(1/2)and f=(cp+k)/(p^2+q^2)^(1/2)

ReplyDeleteRHS = ((a+c)p+bq+3k)/(p^2+q^2)^(1/2) ----(1)

Now G is [(a+c)/3,b/3]Hence, LHS= 3g =3[(a+c)p/3+bq/3+k]/(p^2+q^2)^(1/2)= ((a+c)p+bq+3k)/(p^2+q^2)^(1/2)=RHS by (1). Hence,3g = d + e + f

Ajit: ajitathle@gmail.com

let be M , point BG meet AC

ReplyDeletedraw A'MC' // DF , meet g on G' and e on B'

tr MGG' and MBB' are similar( ang MGG' = ang MBB')

=> (g - (d+f)/2 )/1 = ( e - (d+f)/2)/3 , (1=GM, 3=BM)

2g-d-f = (2e-d-f)/3

6g-3d-3f = 2e-d-f

3g = d+f+e