Sunday, April 9, 2017

Geometry Problem 1328: Two Squares Side by Side, 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 1328: Two Squares Side by Side, Perpendicular, 90 Degrees, Congruence.

Posted by Antonio Gutierrez at 10:29 AM
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6 comments:

  1. APOSTOLIS MANOLOUDISApril 9, 2017 at 12:00 PM

    Problem 1328
    Is AE perpendicular in CG and HK ,then CG//=HK.But <CHG=<HGD=45=<CAG,so
    A,H,C and G are concyclic (AG//HC) .So AHCG is isosceles trapezoid , then
    AH=CG. Therefoore AH=HK.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  2. c.t.e.oApril 9, 2017 at 1:14 PM

    AC bisector of HAE => HAD=45°+HAE, AGC=45°+EGC
    => HAG=AGC (CAE=EGC)

    ReplyDelete
  3. Ignacio Larrosa CañestroApril 9, 2017 at 2:49 PM

    Triangle HAE is symmetrical with respect to AC
    ==> Triangle AKH isósceles in H ===> AH = HK

    ReplyDelete
  4. Peter TranApril 9, 2017 at 5:52 PM

    https://goo.gl/photos/nFqf1YZ178sWAnWn6

    Draw CG and HM ( see sketch)
    Per the result of problem 1327 AE⊥CG => HK//CG
    So HCGK is a parallelogram => HK= CG
    triangle HCE is 45-45-90 triangle so CH= CE
    HB= CB-CH= CD-CE= ED=DG
    Triangle HAM congruent to CDG ( case SAS)
    So ∠ (HAM)= ∠ (CGD)= ∠ (HKM) => AHK is isoceles
    So HA=HK

    ReplyDelete
  5. Sumith PeirisApril 9, 2017 at 7:50 PM

    Reference my proof for Problem 1327 AE and CG are perpendicular as are AE and HK

    Hence HK//CG and so HCGK is a parallelogram.

    So HK = CG

    Now AH = AE since Tr.s AHC and AEC are congruent SAS.

    But AE = CG since Tr.s AED and CGD are congruent SAS.

    So HK = CG = AE = AH

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. AnonymousApril 12, 2017 at 12:55 PM

    Without using the results of problem - 1327
    Drop a perpendicular from H to AD and denote it as M.
    Since m(AJH) = m(AMH) = m(ABH) = 90 => ABHJM are concyclic
    Therefore m(MJH) = m(MAJ) and since HM = AD => triangles AED congruent to HKM (ASA)
    => AE = HK --------- (1)
    Since m(HEC) = 45 => HCE is right angle isosceles and AC bisects HE => AHE is isosceles triangle
    => AH = AE ---------- (2)
    From (1) and (2) => AH = HK

    ReplyDelete

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