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  1. #1
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    Question Use vectors to prove ...

    question 1
    Use vectors to prove that the line segment joining the midpoint of two sides of a triangle is parallel to the third side and half as long.

    Question 2
    Use vectors to prove that the midpoints of the sides of a quadrilateral are the vertices of a parallelogram.

    Can you show me how to do them? Thank you very much.
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  2. #2
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    Quote Originally Posted by Jenny20 View Post
    question 1
    Use vectors to prove that the line segment joining the midpoint of two sides of a triangle is parallel to the third side and half as long.

    Question 2
    Use vectors to prove that the midpoints of the sides of a quadrilateral are the vertices of a parallelogram.

    Can you show me how to do them? Thank you very much.
    Well, two things first,

    1) HAPPY NEW YEAR TO ALL !

    2) My Pittsburgh Steelers have just eliminated the Cinci Bungles from NFL playoffs! WE DEY!

    -----------------
    Okay, for your Question #1 now.

    Here is one way. It's very crude because I'd be using numbers or definite lengths for the vectors. Variables are better for proofs, but hey, it's holiday today. Why crack our heads on a holiday?

    Say we have triangle whose vertices are A(0,0), B(8,6) and C(10,0).
    The two sides AB and AC are halved each. D(4,3) is midpoint of AB. E(5,0) is midpoint of AC.

    b) Is DE half as long as BC?
    DE = sqrt[(4-5)^2 +(3-0)^2] = sqrt[1 +9] = sqrt(10) units long.
    BC = sqrt[(8-10)^2 +(6-0)^2] = sqrt[4 +36] = sqrt(40) = 2sqrt(10) units long.
    Therefore, yes, DE is half in length of BC.-------proven.

    a) Is DE parallel to BC?
    DE = AE -AD -----in vectors.
    DE = ((4 -5),(3-0))
    DE = (-1,3) ---------------***

    BC = AC -AB .....in vectors.
    BC = ((8-10),(6-0))
    BC = (-2,6)
    or,
    BC = 2(-1,3) ---------------***

    Since BC is just DE multiplied by a scalar of 2, then BC and DE are parallel. ----proven.

    ----------------
    That's all for now. I just wanted to say Happy New Year and to needle the Bungleds.
    Last edited by ticbol; December 31st 2006 at 02:08 PM.
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  3. #3
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    Quote Originally Posted by Jenny20 View Post

    Question 2
    Use vectors to prove that the midpoints of the sides of a quadrilateral are the vertices of a parallelogram.

    Can you show me how to do them? Thank you very much.
    Umm, before I lose this, let me prove what's in your Question #2 by words only.

    Take any convex quadrilateral. Draw the two diagonals.

    Any diagonal divides the quadrilateral into two triangles with a common side or base---the diagonal. Draw the line segments connecting the midpoints of the other two sides for each triangle. Remembering your Question #1 above, each of these line segments is half of the diagonal and each is parallel to the diagonal. So both line segments are equal in length and are parallel to each other.

    A little more thinking (like, in a parallelogram, opposite sides are equal and parallel), add two and two, and you have the proof for your Question #2.
    Last edited by ticbol; December 31st 2006 at 02:07 PM.
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  4. #4
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    The above discussion of #2 assumes that the quadrilateral is convex. But it need not be for this to be true. Suppose that ABCD is a quadrilateral and J is midpoint of AB, K is the midpoint of BC, L is the midpoint of CD, and M is the midpoint of DA. Now we prove that JKLM is a parallelogram.
    \begin{array}{rcl}<br />
 \overrightarrow {JK}  & = & \frac{1}{2}\overrightarrow {AB}  + \frac{1}{2}\overrightarrow {BC}  \\ <br />
 \overrightarrow {LM}  & = & \frac{1}{2}\overrightarrow {CD}  + \frac{1}{2}\overrightarrow {DA}  \\ <br />
 \overrightarrow {JK}  + \overrightarrow {LM}  & = & \frac{1}{2}\left( {\overrightarrow {AB}  + \overrightarrow {BC}  + \overrightarrow {CD}  + \overrightarrow {DA} } \right) = 0 \\ <br />
 \overrightarrow {JK}  & = &  - \overrightarrow {LM}  \\ <br />
 \end{array}.

    Thus JKLM has opposite sides that are parallel and have the same length.
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  5. #5
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    Hello, Jenny!

    1) Use vectors to prove that the line segment joining the midpoint of two sides of a triangle
    is parallel to the third side and half as long.
    Code:
                A
                *
               *  *
              *     *
           D *--------* E
            *           *
           *              *
          *-----------------*
          B                   C

    Let D and E be the midpoints of AB and AC, respectively.
    Draw line segment DE.

    We know that: . \overrightarrow{BA} + \overrightarrow{AC} \:=\:\overrightarrow{BC} [1]

    We know that: . \overrightarrow{DA} + \overrightarrow{AE} \:=\:\overrightarrow{DE} [2]

    We are told that: . \overrightarrow{DA} \:=\:\frac{1}{2}\overrightarrow{BA} .and . \overrightarrow{AE} \:=\:\frac{1}{2}\overrightarrow{AC}

    Substitute into [2]: . \overrightarrow{DE} \:=\:\frac{1}{2}\overrightarrow{BA} + \frac{1}{2}\overrightarrow{AC} \:=\:\frac{1}{2}\left(\overrightarrow{BA} + \overrightarrow{AC}\right)

    From [1], we have: . \overrightarrow{DE} \:=\:\frac{1}{2}\overrightarrow{BC}


    Therefore: . \overrightarrow{DE} \,\parallel\, \overrightarrow{BC} .and . \left|\overrightarrow{DE}\right| \:=\:\frac{1}{2}\left|\overrightarrow{BC}\right|

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  6. #6
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    I worked out two proofs for these, but Plato and Soroban beat me. I am going to post anyway, so there.


    #1:

    Let a, b, and c be vectors along the sides of a triangle. and A,B the midpoints of a and b. Then,

    u=\frac{1}{2}a-\frac{1}{2}a=\frac{1}{2}(a-b)=\frac{1}{2}c

    so u is parallel to c and half as long.

    #2:

    Let a,b,c,d be vectors along the sides of the quadrilateral and A,B,C,D be the corresponding midpoints, then

    u=\frac{1}{2}b+\frac{1}{2}c and

    v=\frac{1}{2}d-\frac{1}{2}a

    but d=a+b+c

    so, v=\frac{1}{2}(a+b+c)-\frac{1}{2}a=\frac{1}{2}b+\frac{1}{2}c=u

    Therefore, Hence and moreover, ABCD is a parallelogram because sides AD and BC are equal and parallel.

    Click on the links to see the respective diagrams. There not much, but I hope they help.
    Last edited by galactus; November 24th 2008 at 05:39 AM.
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