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.

Details: Click on the figure below.

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

4 comments:

  1. 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

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  2. 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

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  3. 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

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  4. 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

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