Sunday, October 12, 2008

Elearn Geometry Problem 189

Squares, Distances

See complete Problem 189 at:

Squares, Distances. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. angle between EB and AB equals angle between FE and DA
    construct ∥gram EAOF
    ⊿AOD≡⊿BEA, OD=EA=FO, ∠DOF=∠FEB=90°

  2. with complex numbers
    zD=1; zB=i ; zE= aexp(it)
    if we turn B about E by (-90)we have F
    zF= -i(i-zE)+zE= 1+zE(1+i)
    zDF= zF-zD= zE(1+i)
    the norm from (1+i)is sqr(2), DF=x.sqr(2)

  3. Solution of problem 189.
    The line through E parallel to AD meets AB at H. The line through F parallel to CD meets AD at K, and meets HE at J. We have <FEJ = 90º - <BEH = <EBH, and BE = EF, so triangles BEH and EFJ are congruent. Putting BH = p, HA = q, and HE = r, then EJ = p and FJ = r. Besides AB = AD, so KD = AD – HJ = (p+q) – (p+r) = q – r.
    In triangle AHE, a^2 = q^2 + r^2. In triangle DKF, x2 = (q+r)^2 + (q-r)^2 = 2q^2 +2r^2 = 2a^2. Hence x = a.sqr(2).

  4. <EBD+<DBF=45=<EBD+<ABE, so <ABE=<DBF. Next BF/BE=BD/BA=sqrt2 so triangle DBF is similar to triangle ABE with scale of sqrt2