## Saturday, December 3, 2016

### Geometry Problem 1291 Three Squares, Collinear Vertices, Midpoint, Congruence

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. 1. https://goo.gl/photos/mimSHwEKYejTtxPm9

Let M,N, P, O are the projection of J, E, G and B over AC
Note that triangles CMJ congruent to ENC . Triangle ENA congruent to APG
So CM=EN= u and JM=CN=v ( see sketch)
and EN=AP=u and AN=PG= v1
OP=OA+AP=w+u =OC+CM=OM => O is the midpoint of PM
AC=AN+NC=v1+v=BD=2.w=> w=1/2(v+v1)
In trapezoid GPMJ , middle base OB’=1/2(PG+MJ)= ½(v1+v)=OB
So B coincide to B’ => G, B, J are collinear and B is the midpoint of GJ

2. Prolem 1291
Triangle GAB=triangle AED (AG=AE,AB=AD,<GAB=90+<EAB=<EAD), so GB=ED and <GBA=<EDA.But triangle ECD=triangle JCB (EL=LJ,CD=CB, <ECD=<BCJ=90+<ECB ), so
BJ=DE and <CBJ=<CDE.Then GB=BJ and <GBJ=<GBA+<ABC+<CBJ=90+<ADE+<EDC=90+90=180. Therefore the points G,B and J are
collinear.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

3. It is also true that in this question FB=BH and FB is perpendicular to BH. Any suggestions for proof?

1. 4. Found it: Let P be the midpoint of GE; Q the midpoint of EJ
PEQB is a parallelogram (PB is midpoint segment in Triangle EGJ)
Triangle FPB congruent with Triangle BQH
Hence, FB=BH
<FBP+<FBH+<HBQ+<BQE=180 (cointerior angles in p'gram) I
<HBQ+<BQE+<QHB=90 (adds up to 180 in triangle) II

<FBP=<QHB (matching angles in congruent triangles)

Subtract II from I: <FBH=90

1. Find another way to prove that FB=BH and FB ⊥ BH

See the drawing

5. See the drawing

∠GAB= ∠GAE+∠EAB=Π/2+∠EAB
=> ΔGAB is congruent to ΔEAD (SAS)
=> GB=ED
=> ∠GBA= ∠EDA

In the same way, ΔJCB is congruent to ΔECD (SAS)
=>JB=ED
=> ∠CBJ= ∠EDC

GB=ED and JB=ED Therefore BG=BJ

∠GBJ = ∠GBA +∠ABC+∠CBJ= ∠ADE+Π/2 +∠EDC= Π
Therefore G,B and J are aligned

Define I intersection of GJ and ED
∠CJI= ∠CEI => CJIE are concyclic
=> ∠ECJ= ∠EIJ => ∠EIJ = Π/2
Therefore GJ ⊥DE

From Problem 496
: CF=AH, CF⊥ AH and CF AH and GJ are concurrent (K)
CF⊥ AH => KF⊥ KH => FH diameter of circle FKH

∠AKC= ∠ABC = Π/2
=>ABKC are concyclic
=> ∠BAK= ∠BCK
=> ∠BAH= ∠BCF
Δ BAH is congruent to Δ BCF (SAS)
=> BH=BF and ∠BHA= ∠BFC
=> ∠BHK= ∠BFK
=>FBKH are concyclic
KF⊥ KH Therefore BF⊥ BH

CK⊥ KH and CE⊥ EH =>CKEHJ are concyclic
=> ∠EKC= 3Π/2= ∠EKJ + ∠JKC = ∠EKJ + ∠JEC = ∠EKJ + Π/2
=> ∠EKJ= Π/2
K and I on GJ, IE⊥ GJ and KE⊥ GJ
Therefore K=I