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 = 90ยบ - <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

  5. See the drawing : Drawing

    BGC is congruent to BEA (by rotation of PI/2)
    BA⊥BC, BE⊥BG => AE⊥CG

    Define H such as AHD is the translation of BGC
    AHD is congruent to BGC (by translation)
    => BG=AH=EF and BG//AH//EF
    AE⊥CG, DH//CG => AE⊥DH
    EF=AH and EF//AH => AE=HF and AE//HF
    HF//AE and AE⊥DH => HF⊥DH

    Therefore x^2=2a^2 or x=2√��

  6. Tr. BEA= Tr.BGC. Hence GC=a
    Draw GO perpendicular to BC such that GO=BC
    Connect AO,FO and DO to form three parallelograms AOGB, AOFE and DOGC.
    It can be easily seen that DO=GC=a, FO=AE=a and since m(BEA)=m(AOD) and m(AEF)=m(AOF) => m(FOD)=m(BEF)=90
    Clearly FAD is an isosceles right triangle => x=a.Sqrt(2)