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.

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

Posted by Antonio Gutierrez at 11:52 AM
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7 comments:

  1. Peter TranDecember 3, 2016 at 5:45 PM

    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

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  2. APOSTOLIS MANOLOUDISDecember 3, 2016 at 11:26 PM

    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

    ReplyDelete
  3. Ori GolanJuly 22, 2017 at 5:46 PM

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

    ReplyDelete
  4. AnonymousJuly 23, 2017 at 5:33 AM

    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



    ReplyDelete
    Replies
    1. rv.littlemanDecember 13, 2021 at 11:14 AM

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

      See the drawing

      Delete
  5. rv.littlemanDecember 6, 2021 at 2:14 PM

    See the drawing


    ∠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 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

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