Monday, April 10, 2017

Geometry Problem 1329: Two Squares Side by Side, Parallel, Perpendicular, 90 Degrees, Congruence

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

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Geometry Problem 1329: Two Squares Side by Side, Parallel, Perpendicular, 90 Degrees, Congruence.

Posted by Antonio Gutierrez at 7:48 AM
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7 comments:

  1. APOSTOLIS MANOLOUDISApril 10, 2017 at 9:47 AM

    Problem 1329
    Is CM//EG then ECMG=parallelogram.So CE=MG.But <EHC=<CEH=45=<EGM then
    HC=CE=MG so HCMG is isosceles trapezoid (cyclic) or HM=CG.But the AHCG is isosceles
    Trapezoid (cyclic) and so the points A,H,C,M and G are concyclic.Now <HAM=<HGM=45
    or arc AH=arc HM or AH=HM and <AHM=<AGM=90.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  2. APOSTOLIS MANOLOUDISApril 10, 2017 at 10:20 AM

    Problem 1329 Solution 2
    Suppose that the BC intersects GM in point N then NG=NH=AB (<GHN=45=<HGN=<MCN)
    and CN=NM=DG=BH(AG=BN or AD+DG=BH+HN or DG=BH).Istriangle ABH=triangle HNM
    so AH=HM and <BAH=<MHN but <BAH+<BHA=90 or <BHA+<MHN=90.Therefore
    <AHM=90.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  3. Peter TranApril 10, 2017 at 1:08 PM

    https://goo.gl/photos/rZBQU3ty6Rsrpccq6

    BC meet FG at N
    Observe that HCE and CNM are 45-45-90 triangles
    So HB=ED=CN=DG
    Triangles ABH and HNM are congruent => AH=HM and angle AHM= 90

    ReplyDelete
  4. AnonymousApril 13, 2017 at 6:57 AM

    Since HCE is right-angled isosceles, => BH=ED
    Also CM||EG => MG=CE => HCMG is an isosceles trepezoid
    Join CG and let m(GCD)=x => m(GCM) = 45-x = m(GHM) -----------(1)
    Drop a perpendicular from H to meet AG at P and since HPG is isosceles triangle => m(PHG) = 45 --------(2)
    Also the triangles ABH and CDG are congruent (SAS) => m(BHA) = 90-x => m(AHP) = x -------------(3)
    Therefore m(AHM) = m(AHP)+m(PHG)+m(GHM) = 90
    Since m(AGM) = m(AHM) = 90 => AHMG are concyclic
    Join AM and since m(MGH) = 45 => m(MAH) = 45 and the triangle AHM is right-angled isosceles => AH=HM

    ReplyDelete
  5. AnonymousJune 29, 2017 at 10:50 AM

    To prove AH=HM:
    Let intersection point of CM and EF be point I.
    <HGM+<GHM = <HMC+<CMF (ext.<)
    Since <HGM=45° and <HMC=<GHM
    <CMF=45°
    By using this result, it can be deduced that triangle HCE,CEI and IFM are isosceles.

    Let length of side of large square be 1.
    BH^2+1^2=AH^2
    (HC+EF)^2+(CE+FM)^2=HM^2 (both Pyth.)

    BH+HC=CE+ED=1
    HC=CE (isos.)
    So BH=ED
    HC+ED=1
    Since ED=EF,
    HC+EF=1

    CE=EI (isos.)
    IF=FM (isos.)
    EI+IF=EF
    Hence CE+FM=1

    ED^2+1^2=AH^2
    1^2+ ED^2=HM^2 (both identical)
    Hence AH=HM

    ReplyDelete
  6. AnonymousJune 29, 2017 at 10:59 AM

    To solve <AHM=90°,
    construct point J so that CDGJ is a rectangle and BC extended meets J.
    Triangle ABH is congruent to HMJ
    <BAH=<JHM
    <BAH+<BHA+<90°=180°
    <JHM+<BHA+<AHM=180°
    Hence <AHM=90°

    ReplyDelete
  7. Sumith PeirisSeptember 8, 2023 at 2:45 AM

    Let the larger square be of side a and the smaller square be of side b
    Complete Rectangle CEFN

    Triangle HNG is Right Isosceles so HN = a
    Triangle CNM is Right Isosceles so NM = b

    Further BH = (a+b) - a = b

    So Triangles AGH & GNH are congruent SAS (AB = HN = a, BH = NM = b & the included angle is 90)

    Hence HM = AH

    Also < AHM = 90 since < AHG & < MHN (= < GAH) are complementary

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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