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Math Help - Difficult Similarity Proof

  1. #1
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    Difficult Similarity Proof

    Prove that any angle bisector of a triangle separates the opposite side into segments whose lengths have the same ratio as the ratio of the lengths of the remaining two sides.

    I know that I need to construct a line parallel to a certain line segment of the triangle to from similar triangles... From here I am lost...
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  2. #2
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    Hello, MATNTRNG!

    Here is one form of the proof . . .


    Prove that a angle bisector of a triangle separates the opposite side into segments
    whose lengths have the same ratio as the ratio of the lengths of the other two sides.

    We have \Delta ABC with angle bisector AD.
    Let: . \angle BAD = \angle CAD = \theta
    Let: . AB = c,\;AC = b

    Code:
                        B
                        *
                     *  :*
              c   *     : *
               *       h:  *
            *           :   *
         *  θ           :    * D          F
    A * . - - - - - - - + - - o - - - - - + 
         *  θ           E      *          :
            *                   *         :
               *                 *        :
                  *               *       :
                   b *             *      :k
                        *           *     :
                           *         *    :
                              *       *   :
                                 *     *  :
                                    *   * :
                                       * *:
                                          o
                                          C

    Draw h = BE perpendicular to AD.
    Draw k = CF perpendicular to AD.

    Since \Delta BED \sim \Delta CFD\!:\;\;\dfrac{BD}{DC} \,=\,\dfrac{h}{k} .[1]


    In right triangle BEA\!:\;\sin\theta \,=\,\dfrac{h}{c} \quad\Rightarrow\quad h \:=\:c\sin\theta

    In right triangle CFA\!:\;\sin\theta \,=\,\dfrac{k}{b} \quad\Rightarrow\quad k \,=\,b\sin\theta


    Substitute into [1]:. \dfrac{BD}{DC} \:=\:\dfrac{c\sin\theta}{b\sin\theta}


    Therefore: . \dfrac{BD}{DC} \:=\:\dfrac{c}{b}

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  3. #3
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    difficult similarity problem

    Follows a solution not using trig

    draw an isosceles triangle XYZ apex @X. draw XK perpendicular toYZ.from Y draw any line toXZ meeting it @M and perpen @P
    draw a line from M parallel to YZeeting perpen @S

    label XY=a XM=b YP=c PM=d
    triangleMSP sim to YPK therefor c/d= YK/MS

    triangle XSM sim to XYK therefor a/b= YK/MS

    a/b= c/d not mentioned above XK bisects angle X

    bjh
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  4. #4
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    Quote Originally Posted by bjhopper View Post
    Follows a solution not using trig

    draw an isosceles triangle XYZ apex @X. draw XK perpendicular toYZ.from Y draw any line toXZ meeting it @M and perpen @P
    draw a line from M parallel to YZeeting perpen @S

    label XY=a XM=b YP=c PM=d
    triangleMSP sim to YPK therefor c/d= YK/MS

    triangle XSM sim to XYK therefor a/b= YK/MS

    a/b= c/d not mentioned above XK bisects angle X

    bjh
    Wouldnt this proof only apply to isoceles triangles tho???
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  5. #5
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    diffcult similarity problem

    [QUOTE=MATNTRNG;535172]Wouldnt this proof only apply to isoceles triangles tho???[/QUOT

    this proof applies to any triangle.draw any triangle and bisect one of its angles. extend the shorter side of this angle so that the two legs are the same lenght. connect the now equal sides.this is the created isosceles triangle.your triangle is still intact.


    bjh
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  6. #6
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    Quote Originally Posted by bjhopper View Post
    Follows a solution not using trig

    draw an isosceles triangle XYZ apex @X. draw XK perpendicular toYZ.from Y draw any line toXZ meeting it @M and perpen @P
    draw a line from M parallel to YZeeting perpen @S

    label XY=a XM=b YP=c PM=d
    triangleMSP sim to YPK therefor c/d= YK/MS

    triangle XSM sim to XYK therefor a/b= YK/MS

    a/b= c/d not mentioned above XK bisects angle X

    bjh
    I am sorry but I am having a great deal of difficulty constructing your proof. I am following your line of resoning with constructing the isoceles triangle but its just not coming together for me. I cant see how triangle XSM is similar to XYK. I must be doing something wrong. I would really like to know how to prove this without trig so if u could dumb down the proof for me it would be greatly appreciated. Thank you
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  7. #7
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    If you could provide a picture that would be great too. Thanks
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  8. #8
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    Hello MATNTRING,
    I am sorry that you do not grasp this.Partly my fault because originally did not say that the solution requires construction lines.These lines give a figure in which the a,b,c,d are related.In my last post I described the construction lines.If you do as I said you will have a figure consisting of any given triangle and an added section which forms together with the original triangle an isosceles triangle.I have labeled this figure.

    YPK is sim to MSP because angle SMP =PYK and YPK=MPS

    XSMis a right tri and XYK =rt tri Both tris have equal angles at X becauseXK bisects X


    Does this help?


    bjh
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