## Sunday, October 12, 2008

### Elearn Geometry Problem 189

See complete Problem 189 at:
www.gogeometry.com/problem/p189_square_fact.htm

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

1. angle between EB and AB equals angle between FE and DA
construct ∥gram EAOF
⊿AOD≡⊿BEA, OD＝EA＝FO, ∠DOF＝∠FEB＝90°
x＝OF√2＝a√2

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)