Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Saturday, December 3, 2016

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

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https://goo.gl/photos/mimSHwEKYejTtxPm9

ReplyDeleteLet 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

Prolem 1291

ReplyDeleteTriangle 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

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

ReplyDeleteSuggestions: See the problem 496 at

Deletehttp://www.gogeometry.com/problem/p496_triangle_two_square_side_concurrent_90_degrees.htm

Found it: Let P be the midpoint of GE; Q the midpoint of EJ

ReplyDeletePEQB 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

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

DeleteSee the

drawingSee the

ReplyDeletedrawing∠GAB= ∠GAE+∠EAB=Π/2+∠EAB

∠EAD= ∠EAB+ ∠BAD=∠EAB+ Π/2

=> ∠GAB= ∠EAD

=> 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∠ADC= Π/2=∠ADE+∠EDC

∠GBJ = ∠GBA +∠ABC+∠CBJ= ∠ADE+Π/2 +∠EDC= Π

Therefore G,B and J are alignedDefine I intersection of GJ and ED

∠CJI= ∠CEI => CJIE are concyclic

=> ∠ECJ= ∠EIJ => ∠EIJ = Π/2

Therefore GJ ⊥DEFrom

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⊥ BHCK⊥ 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