Wednesday, January 21, 2009

Elearn Geometry Problem 229: Triangle, Centroid, Transversal

Triangle, Midpoints, Exterior line

See complete Problem 229 at:

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. 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)
    or 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

  2. like P227
    draw 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