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Math Help - Plz Help me with this triangle proof

  1. #1
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    Plz Help me with this triangle proof

    Prove analytically that the line joining the middle points of two sides of a triangle is parallel to the third side and equal half of its length.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by the_rookie07 View Post
    Prove analytically that the line joining the middle points of two sides of a triangle is parallel to the third side and equal half of its length.
    Let A, B, C be the vertices of the triangle plotted on the Cartesian Coordinate plane, so let

    A= (x_a,y_a)
    B= (x_b,y_b)
    C= (x_c,y_c)

    lets take the midpoints of BC and AC

    BC_m = \left( {\frac{x_b + x_c}{2}, \frac{y_b + y_c}{2}} \right)

    AC_m = \left( {\frac{x_a + x_c}{2}, \frac{y_a + y_c}{2}} \right)

    using two-point form, the line joining the two midpoints is

    y - \frac{y_a + y_c}{2} = \frac{ \frac{y_b + y_c}{2} - \frac{y_a + y_c}{2} } {\frac{x_b + x_c}{2} - \frac{x_a + x_c}{2}} \left( {x - \frac{x_a + x_c}{2}} \right) = \frac{y_b + y_c - y_a - y_c}{x_b + x_c - x_a - x_c} \left( {x - \frac{x_a + x_c}{2}} \right)

    \implies y - \frac{y_a + y_c}{2} = \frac{y_b - y_a}{x_b - x_a } \left( {x - \frac{x_a + x_c}{2}} \right)

    but what is the formula of the line joining AB?

    y - y_a = \frac{y_b - y_a}{x_b - x_a} \left( {x - x_a} \right)

    and they have equal values of slopes.. hence, they are parallel.. QED
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  3. #3
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    hey thanks alot Kalagota...
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  4. #4
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    Quote Originally Posted by the_rookie07 View Post
    Prove analytically that the line joining the middle points of two sides of a triangle is parallel to the third side and equal half of its length.
    Here is one way.

    Draw any triangle ABC.
    D and E are midpoints of AB and BC, respectively.
    Let AB = x; BC = y; CA = z; DE = w

    To prove analytically that z = 2w. Using Law of Cosines:

    In triangle ABC,
    z^2 = x^2 +y^2 -2(x)(y)cosB
    cosB = (x^2 +y^2 -z^2) /(2xy) -------**
    In triangle DBE,
    w^2 = (x/2)^2 +(y/2)^2 -2(x/2)(y/2)cosB
    cosB = [(1/4)x^2 +(1/4)y^2 -w^2] /[(1/2)(xy)] ------**

    cosB = cosB,
    (x^2 +y^2 -z^2) /(2xy) = [(1/4)x^2 +(1/4)y^2 -w^2] /[(1/2)(xy)]
    Clear the fractions, multiply both sides by 2xy,
    x^2 +y^2 -z^2 = (2/(1/2))[(1/4)x^2 +(1/4)y^2 -w^2]
    x^2 +y^2 -z^2 = (4)[(1/4)x^2 +(1/4)y^2 -w^2]
    x^2 +y^2 -z^2 = x^2 +y^2 -4w^2]
    -z^2 = -4w^2
    z^2 = 4w^2
    Therefore, z = 2w ---------proven.

    ---------------------------------------------
    To prove that DE is parallel to AC. Using Law of Cosines again,

    In triangle ABC,
    y^2 = x^2 +z^2 -2(x)(z)cosA
    cosA = (x^2 +z^2 -y^2) /(2xz) -------**

    In triangle DBE,
    (y/2)^2 = (x/2)^2 +w^2 -2(x/2)(w)cosD
    cosD = [(1/4)x^2 +(z/2)^2 -(1/4)y^2] /[2(x/2)(z/2)]
    cosD = [(1/4)x^2 +(1/4)z^2 -(1/4)y^2] /[2(1/4)xz]
    cosD = [x^2 +z^2 -y^2] /[2xz] -----------------------**

    So, cosA = cosD
    Hence, angle BAC = angle BDE
    Since BD is parallel to BA, then, DE is parallel to AC. -------proven.
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  5. #5
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by ticbol View Post
    Here is one way.

    Draw any triangle ABC.
    D and E are midpoints of AB and BC, respectively.
    Let AB = x; BC = y; CA = z; DE = w

    To prove analytically that z = 2w. Using Law of Cosines:

    In triangle ABC,
    z^2 = x^2 +y^2 -2(x)(y)cosB
    cosB = (x^2 +y^2 -z^2) /(2xy) -------**
    In triangle DBE,
    w^2 = (x/2)^2 +(y/2)^2 -2(x/2)(y/2)cosB
    cosB = [(1/4)x^2 +(1/4)y^2 -w^2] /[(1/2)(xy)] ------**

    cosB = cosB,
    (x^2 +y^2 -z^2) /(2xy) = [(1/4)x^2 +(1/4)y^2 -w^2] /[(1/2)(xy)]
    Clear the fractions, multiply both sides by 2xy,
    x^2 +y^2 -z^2 = (2/(1/2))[(1/4)x^2 +(1/4)y^2 -w^2]
    x^2 +y^2 -z^2 = (4)[(1/4)x^2 +(1/4)y^2 -w^2]
    x^2 +y^2 -z^2 = x^2 +y^2 -4w^2]
    -z^2 = -4w^2
    z^2 = 4w^2
    Therefore, z = 2w ---------proven.

    ---------------------------------------------
    To prove that DE is parallel to AC. Using Law of Cosines again,

    In triangle ABC,
    y^2 = x^2 +z^2 -2(x)(z)cosA
    cosA = (x^2 +z^2 -y^2) /(2xz) -------**

    In triangle DBE,
    (y/2)^2 = (x/2)^2 +w^2 -2(x/2)(w)cosD
    cosD = [(1/4)x^2 +(z/2)^2 -(1/4)y^2] /[2(x/2)(z/2)]
    cosD = [(1/4)x^2 +(1/4)z^2 -(1/4)y^2] /[2(1/4)xz]
    cosD = [x^2 +z^2 -y^2] /[2xz] -----------------------**

    So, cosA = cosD
    Hence, angle BAC = angle BDE
    Since BD is parallel to BA, then, DE is parallel to AC. -------proven.
    this is a nice proof!
    but i assume s/he doesn't know what laws of cosine are yet..
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  6. #6
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by the_rookie07 View Post
    hey thanks alot Kalagota...
    i forgot how to show the distance thing..
    i assume you know what to do.. (just use distance formula on A and B and on the midpoints..)
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  7. #7
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    hey guys you have been of great help! thanks alot...u guys are like superheroes saving the world at one geometric problem at a time...thanks
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  8. #8
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    Hello, the_rookie07!

    Prove analytically that the line joining the midpoints of two sides of a triangle
    is parallel to the third side and equal half of its length.

    Given any triangle, we can orient it so that one vertex is at the origin
    . . and another is on the x-axis.
    Code:
              |
              |           B
              |           o(p,q)
              |         *  *
              |       *     *
              |   D o--------o E
              |   *           *
              | *              *
          - A 0 * * * * * * * * o C -
            (0,0)             (r,0)
              |

    We have the vertices: . A(0,\,0),\;B(p.\,q),\;C(r,\,0)

    The midpoint of AB is: . D\left(\frac{p}{2},\:\frac{q}{2}\right)

    The midpont of BC is: . E\left(\frac{p+r}{2},\:\frac{q}{2}\right)


    The slope of AC is: . m_{_{AC}} \:=\:\frac{0-0}{r-0} \;=\;0

    The slope of DE is: . m_{_{DE}} \:=\:\frac{\frac{q}{2} - \frac{q}{2}}{\frac{p+r}{2} - \frac{p}{2}} \;=\;0

    . . Therefore: . {\color{blue}DE \parallel AC}


    The length of DE \:=\:\sqrt{\left(\frac{p+r}{2} - \frac{p}{2}\right)^2 + \left(\frac{q}{2} - \frac{q}{2}\right)^2} \;=\;\sqrt{\left(\frac{r}{2}\right)^2} \;=\;\frac{r}{2}

    The length of AC \:=\:r

    . . Therefore: . {\color{blue}DE \:=\:\frac{1}{2}AC}

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  9. #9
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    hey thanks Soroban, nice n simple u make that look easy
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  10. #10
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    Hi Rookie, there is an easier solution I think using vectors.

    Use soroban's triangle picture and imagine you are walking these journeys.

    You can see that:-

    de = db + be

    But also:-

    ac = ad + db + be + ec

    But point d bisects ab and point e bisects bc therefore:-
    ad = bd
    and
    be = ec

    thus, ac = 2db + 2be
    or:-
    ac = 2( db + be)

    but we already stated that de = db + be
    so,

    ac = 2bd
    therefore:-
    bd = 1/2 ac.

    I doubt there's an easier way to prove it.
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