Saturday, October 17, 2009

Problem 367. Square, Angle, 45 degrees, Pythagoras, Congruence, Similarity

Proposed Problem
Click the figure below to see the complete problem 367 about Square, Angle, 45 degrees, Pythagoras, Congruence, Similarity.

Problem 367. Square, Angle, 45 degrees, Pythagoras, Congruence, Similarity.
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Complete Geometry Problem 367
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 9:26 AM
Labels: , , , , ,

8 comments:

  1. muharebOctober 22, 2009 at 2:49 AM

    AB=BC=CD=AD=(a+b+x)/sqrt(2)...(*)

    from similarity between triangles AGD and HGA:
    AG=sqrt(x^2+xb)...(**)

    from similarity between triangles ABH and GAH:

    AB/GA=AH/GH=BH/AH

    by setting the values from (*) and (**); we get:
    AH=x*[(a+b+x)/sqrt(2)]/sqrt(x^2+bx)...(1).
    and
    AH=sqrt(x^2+xa)...(2).

    after solving the equations (1),(2) we get:

    x^2=a^2+b^2

    ReplyDelete
  2. AnonymousNovember 13, 2009 at 4:42 PM

    http://geometri-problemleri.blogspot.com/2009/11/problem-47-ve-cozumu.html

    ReplyDelete
  3. bae deok rakOctober 1, 2010 at 11:40 PM

    since angle(GAF)=angle(GDF)=45 and angle(ADF)=90, GADF is a cyclic and so EGHF is an also cyclic. Hence two triangles BEG and BHC are similar and hence a(a+x)=BE(a+b+x)/sqrt(2).
    Since BE=(a+b+x)/{sqrt(2)(x+b)}, we have
    get x^2 =a^2 +b^2.

    ReplyDelete
  4. AnonymousJanuary 9, 2013 at 7:42 PM

    There is a very simple and elegant solution if we reflect triangle ABE across AE so B becomes B' and we reflect triangle ADF across AF so that D becomes D', then since AB=AD we must have B'=D'. Also angle GB'H=angle GB'A+ angle HB'A=angleGBA+ angle HDA=45+45+90, so by Pythag on triangle GB'H, a^2+b^2=x^2.

    ReplyDelete
  5. Michael TsourakakisJanuary 15, 2013 at 6:10 AM

    Because angle EAH=angle HBE=45, the quadrilateral BAHE is, cyclic quadrilateral.So, The EH is perpendicular to AF.
    Because angle GDF=angle GAH=45, the quadrilateral AGFD is, cyclic quadrilateral.So, the FG is perpendicular to AE. Therefore, The EH, FG, is heights of triangle EAF, so, ALK, is the other height of the triangle AEF
    By cyclic quadrilaterals arising, we have, angle EKG=angle FKH=45,so, angle GKH=90.
    More , angle AEF= angle GHA=angle BEA and angle KFH=angle AGD=angle AFD .
    Therefore, triangle BEG=triangle GEK and triangle KHF=triangle FDH.So ,GK=GB=a and KH=HD=b
    Pythagoras theorem to the triangle GKH: x^2=a^2+b^2
    Image: http://img267.imageshack.us/img267/2807/geogebra.png

    ReplyDelete
  6. UnknownMay 4, 2014 at 1:29 AM

    Tr.GAH & tr.ABH are similar(<GAH=<ABH=45,<H is common).so we have,AG/AB=AH/a+x.......I)
    Tr.GAH & tr.ADG are similar(<GAH=<ADG=45,<G is common).so we have AH/AD=AG/b+x......II)
    multiplying the eq.s,we get AG*AH/AB*AD=AG*AH/(a+x)(b+x)
    therefore,(a+x)(b+x)=AB*AD=c^2[let AB=AD=c]
    therefore,2ab+2ax+2bx+2x^2=2c^2=(c*sqrt.2)^2=BD^2=(a+x+b)^2
    therefore,2x^2=a^2+x^2+b^2
    so we conclude, x^2=a^2+b^2

    ReplyDelete
  7. Sumith PeirisJuly 5, 2015 at 9:35 PM

    Find M such that DM = a and < ADM = 45.
    Tr.s ADM & AEB are congruent from which we deduce that Tr.s AGH & AMH are congruent as well both cases SAS.
    Hence FM = x and applying Pythagoras to Tr. HDM the result follows.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  8. rv.littlemanNovember 15, 2020 at 9:45 AM

    See the drawing

    Define H’ rotation of H with angle –90° and center A
    =>AH=AH’ and BH’=DH=b
    => BH’ ⊥DH <=> BH’ ⊥BG
    ∠HAG=45° and ∠HAH’=90° => ∠GAH’=45°
    ΔGAH is congruent to ΔGAH’ (SAS)
    => GH’=GH=x
    BH’=b, BG=a, BH’ ⊥BG and GH’=x => ΔGBH’ is a rectangle triangle in B
    Therefore x^2=a^2+b^2




    ReplyDelete

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