Monday, May 9, 2016

Geometry Problem 1211: Right Triangle, Altitude, Angle Bisector, 45 Degrees

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

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click the figure below to view more details of problem 1211.


Geometry Problem 1211: Right Triangle, Altitude, Angle Bisector, 45 Degrees

5 comments:

  1. Problem 1211
    Is <EDB+<BDF=45+45=90, <EBF+<EDF=90+90=180,then B,E,D,F are concyclic.Then BE=BF=x
    (<BEF=<BDF=45=<BFE).Therefore EF=x√2 . By the theorem of Ptolemy is BD.EF=DF.x+DE.x=
    X(DF+DE) or DF+DE=(BD.EF)/x=BD. √2.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORRYDALLOS GREECE

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  2. http://s32.postimg.org/eng98hm51/pro_1121.png

    Let G an are the projection of E over BD and F over DC ( see sketch)
    Triangle BED similar to tri. CFD ( case AA)
    So EG/FH= BD/DC= FH/HC= EG/BG => BG= FH
    So EG+ FH= DG+BG= BD
    Replace EG= ED. sqrt(2) and BG=FH=DF.sqrt(2)
    We get ED+DF= DG+FH=BD.sqrt(2)

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  3. Let L be the line through B parallel to AC.
    Extend DF to meet L at G.
    Triangle BDG is right angled and isosceles (BD = BG)
    So DG = BD√2.
    It suffices to prove DE = FG.
    Compare triangles BED and BFG.
    Angle EBD = Angle FBG (each = C).
    Angle EDB = Angle BGF (each = 45 degrees).
    Side BD = Side BG.
    By ASA the above triangles are congruent.
    So DE = FG as required.

    N Vijaya Prasad,
    Rajahmundry,
    INDIA.

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  4. In Quadrilateral BEDF, B+D = 90 and E+F = 135-c+45-c = 90 degrees. So it is cyclic.

    Join EF, since B,E,D, and F are concyclic, BDF = 45 AND FEB =45 degrees. Hence triangle BEF is right isosceles triangle => BE=BF=(1/Sqrt 2)EF -----> (1)

    Since triangle DEF is right angle => Square(BE)=0.5(Square ED + Square FD) -----> (2)

    Triangles EDB & FDC are similar
    => DC = BD(FD/ED) ----- > (3)
    & CF = BE(FD/ED) -------> (4)

    Triangles AED & BFD are similar
    => AD = BD(ED/FD) -------> (5)
    & AE = BF(ED/FD) = BE(ED/FD) (From (1)) --------> (6)

    From (3) & (5) AC = AD+DC = BD.(Square ED + Square FD)/(ED.FD) --------> (7)
    From (1) & (4) BC = BF+CF = BE + BE(FD/ED) = BE (FD+ED)/FD ---------> (8)
    From (1) & (6) AB = BE + AE = BE(FD+ED)/FD -------> (9)

    Applying Pythogrous to Triangle ABC => Square AC = Square AB + Square BC
    => Square BD. Square(Square ED + Square FD)/Square(ED.FD) = Square(BE).Square(FD+ED)/Square(FD) + Square(BE).Square(FD+ED)/Square(ED)

    Replacing BE = 0.5(Square ED + Square FD) from (2) in above equation and solving

    => FD + ED = Sqrt(2) BD

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  5. ∠ADE+∠EDB=∠ADB
    2∠EDB=90
    ∠EDB=45
    ∠BDF=45
    ∠EDF=∠EDB+∠BDF=45+45=90
    ∠EBF+∠EDF=90+90=180
    E,B,F,D concyclic
    ∠BEF=∠BDF=45
    ∠BEF+∠EFB+∠FBE=180
    45+∠EFB+90=180
    ∠EFB=45
    ∠BEF=∠EFB
    BE=BF
    ∠ADB=∠ABC=90
    ∠BAD=∠CAB
    BAD~CAB
    BD/DA=AB/BC
    AE/EB=c/a
    EA=cBE/a
    BE+EA=BA
    BE+cBE/a=c
    aBE+cBE=ac
    BE=ac/(a+c)
    BF=BE
    EF^2=BE^2+BF^2=2BE^2=2a^2
    BD/BA=CB/AC
    BD/c=a/√(a^2+c^2)
    BD=ac/√(a^2+c^2)
    c^2/(a+c)^2
    (ED+DF)^2-EF^2=2BD^2-2a^2c^2/(a+c)^2
    ED^2+2EDDF+DF^2-EF^2=2a^2c^2/
    (a^2+c^2)-2a^2c^2/(a+c)^2
    2EDDF=[(a+c)^2—(a^2+c^2)]2a^2c^2/(a^2+c^2)(a+c)^2=2ac2a^2c^2/((a^2+c^2)(a+c)^2)=4a^3c^3/((a^2+c^2)(a+c)^2)
    Area of DEF=1/2EDDF=1/4(2EDDF)=1/4(4a^3c^3/(a^2+c^2)(a+c)^2)=a^3c^3/((a^2+c^2)(a+c)^2)

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