See complete Problem 229 at:

gogeometry.com/problem/p229_triangle_centroid_transversal_perpendicular.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Wednesday, January 21, 2009

### Elearn Geometry Problem 229: Triangle, Centroid, Transversal

Labels:
centroid,
distance,
perpendicular,
transversal,
trapezoid,
triangle

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Let A be (0,0), B:(a.b) and C:(c,0) Hence, G:((a+c)/3,b/3). Let DF be px+qy+k=0. By convention, distances above this line are considered positive and those below negative. Thus, e = (ap+bq+k)/(p^2+q^2)^(1/2), d=-k/(p^2+q^2)^(1/2), f = -cp/(p^2+q^2)^(1/2) and g= -((a+c)p/3+bq/3 +k)/(p^2+q^2)^(1/2). RHS = -k/(p^2+q^2)^(1/2)-(cp+k)/(p^2+q^2)^(1/2)-(ap+bq+k)/(p^2+q^2)^(1/2)

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

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) --(2)

By (1) & (2) LHS=RHS or 3g = d + f - e

Ajit: ajitathle@gmail.com

like P227

ReplyDeletedraw A'MC'// DF ( M middle of AC)

tr MGG' and MBB" are similar

((d+f)/2-g)/1 = ((d+f)/2+e)/3 give the result