Friday, February 15, 2013

Problem 859: Isosceles Right Triangle, Angles,18, 45 Degrees, Congruence, Perpendicular bisector

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Problem submitted by Charles T.
Click the figure below to see the complete problem 859.

Online Geometry Problem 859: Isosceles Right Triangle, Angles,18, 45 Degrees, Congruence, Perpendicular bisector


  1. Answer: 27
    Sorry for using trigonometry but i can't find a 'normal' solution.
    Let AB=BC=DE=AD=DF=2
    Draw a height BG from B to AC. It is clear from the figure that <DBG=126, <BGE=135 and BG=root(2). From sine theorem BD=4*sin9 and DE=2.
    From cosine theorem(I leave to the reader to use this theorem in triangles DBG and DEG)we see that DEBG is a inscribed quadrilateral.
    So <EDB=45 and <DEF=54. So x=180-2*54-45=27 degrees.

  2. Let is ,AD=DE=DF=AB=BC=α and c , the circle (A,α).
    The line DE , meets the circle c , in point H and the line AE meets the HC in point P. Then , AH = α
    The line EF (perpendicular bisector of BC) meets AC in point M , midpoint of AC and E is midpoint of AP.
    So , frome the triangle ACP we have ,EM//PC . Therefore , HC //AB . So the quadrilateral ABCH we have : ΗΑ=ΑΒ=ΒC=α , HC , AB , perpendicular to BC , so, ABCH is a square.( The proof is easy)
    From the isosceles triangle DAH , angle DAH = 90+18=108 degrees , so, angle DAH=36 degrees. But , from the isosceles triangle ADB , angle ADB=81 degrees ,therefore, angle LDB=81-36=45 degrees.
    From the triangle ADL , angle ALE =18+36= 54 degrees . Because AL // EF, we have , angle LEF=54 degrees .So , from the isosceles triangle DEF , we have, 45+X+54+54=180 , therefore
    X=27 degrees .
    Plan :

    1. Michael

      It is not clear how EM//PC from line 3 and 4 of your solution :"The line EF (perpendicular bisector of BC) meets AC in point M , midpoint of AC and E is midpoint of AP"
      Please provide more details.
      Peter Tran

    2. Hello Peter.You are right.
      hasty my solution.
      I give you one of the two solutions that I've done
      Let is ABCB’ a square. Then , FE is perpendicular bisector of AB’. So AE=EB’=x.
      If, AD=AB=DE=DF=R and c=(A,R) , c’=(D,R) two circles ,then for circle c’ we have
      R^2-AE^2=EB’.DE , R^2-x^2=x.R .Therefore, x=R[(sqrt(5)-1]/2
      So ,for circle c’ , with radius R, the chord AE=x length is R[(sqrt(5)-1]/2 .But this is the chord length of a regular decagon .So , the central angle ADE=36 degrees.
      Then , angle ALE =18+36=54 degrees .But AB//FE ,so, angle LEF = angle DFE =54 degrees.
      Additional ,for isosceles triangle ADB, is, angle ADB=81,therefore, angle LDB=45 degrees.
      Finality ,for triangle EDF 54+54+45+X=180 , so, x=27 degrees
      I hope I have not accidentally


  3. The correct address for the image

  4. prof radu ion,Sc.Bozioru,BuzauFebruary 20, 2013 at 3:43 AM

    {P}= cercul (D,AD) taie pe AC=> <EDA=y
    { T}= BA cu DE,<BTD=18+x
    BA||EF, <BTD=<DEF <EDF=144-2y
    x==144-2y-(81-y)=63-y unde 0<y<54

  5. I extended DB until it intersects EF. I call that point G. I note that DG=DB + BG. I determine that <DGF=99deg. I note that sin81deg = sin99deg. I use the law of sines and substitution to conclude that <F=54 deg. <X = 180 - 99 - 54 = 27 deg.

  6. I can scan my solution with drawing to anyone who is interested. All that is required is to extend line DB until it intersects with line EF to form 2 other triangles. Then use the law of sines and algebraic substitution.

  7. Let AB = BC = AD = DE = DF = a
    Draw DMN perpendicular to AB (or EF)
    meeting AB at M and EF at N.
    Note <ABD = <ADB = 81 deg.
    So <MDB = 90 – 81 = 9 deg.
    From the isosceles ΔABD,
    we have BD = 2a sin 9
    Follows DM = BD cos 9
    = 2a sin 9 cos 9 = a sin 18.
    EF bisects BC and is pependicular to it.
    So MN = a/2.
    So DN = DM + MN = a (sin 18 + 1/2)
    = a[(sqrt(5) – 1)/4) + 1/2]
    = a (sqrt(5) + 1)/4 = a cos 36.
    DN bisects <EDF. So <NDF = x + 9
    and DN = a cos (x + 9).
    Hence a cos (x + 9) = a cos 36,
    cos (x + 9) = cos 36,
    x + 9 = 36,
    x = 27 degrees

  8. aceasta e solutia corecta ,ceruta
    {P}= cercul (D,AD) taie pe AC=> <EDA=y
    { T}= BA cu DE,<BTD=18+y
    BA||EF, <BTD=<DEF <EDF=144-2y
    x==144-2y-(81-y)=63-y unde 0<y<54

    1. To Prof Radu Please review your solution. In this problem we don't need to know "y" to determine "x". Thanks.

    2. Valorile lui x sunt in functie de valorile lui y,unde 0x=27
      daca y=30 =>x=33 etc. In problema punctul E nu e precizat el se poate deplaseasa pe arcul de cerc AP
      {P}= cercul (D,AD) taie pe AC al acestui cerc.Va rog sa verificati.

  9. Daca y=36,atunci x=27 Daca y=30 =>x=33 etc.

    unde 0<y<54,<EDA=y.

  10. In Figure
    since <BHD=9 and <EMH=45, then <DLB=<DEM=54 and <ADL=36. Hence <LDB=45.
    Since DE=DF, and so <DEF=<DFE=54, <EDF=72.
    Furthermore, since <BDL=45, <LDF=72, we see that x=27.

  11. Michael Tsourakakis and 배덕락: Why are D, E, B1, collinear?
    If you cannot prove that it is true, the problem is not solved as you claim.


  13. Let EF,BP,AE respectively meet AC,DE,QC at P,Q,R.

    EF is the perpendicular bisector of BC, hence AP = PC so AE = ER which
    implies that EF // QC.

    Hence < QCR = 45 and so PQ = PC = PB = PA and so ABCQ is a square.

    Therefore < DAQ = 108, < AQD = ADQ = 36 and so < DEN = 18+36 = 54 = < DFE.

    Hence < EDF = 72 and finally x = 72 – (81-36) = 27

    Sumith Peiris
    Sri Lanka