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Midpoint proof

  1. Jul 18, 2004 #1
    How would you right a proof for this theorem: "If a segment is given, then it has exactly one midpoint"?
    Please note that the numbering of the postulates (P) is based on my geometry book. Also, I'm just in 9th grade geometry, so please don't use differential equations or some other math than basic Euclidean geomtery. This will help me better understand the concept.

    This is what I did so far:
    This is where I kind of get lost....

    Does that prove Q is the midpoint? I don't think so because:
    1.) I haven't shown that the arithmetic I did in an attempt to show the distances between AQ and BQ are equal to that of AB works for all cases.

    2.)I'm not sure I have adequately proven that A, Q and B are on the same line.
  2. jcsd
  3. Jul 18, 2004 #2
    Here is the proof my math tutor wrote:

  4. Jul 18, 2004 #3
    Midpoint - Point B is the midpoint of the segment AC if it is between A and C and if AB=BC(that is, the distance from B to A is the same distance from B to C).

    Simply this


    B is located at the midpoint. There is equal distance between A and B, as well there is equal distance from B to C. This shows that B is the midpoint of AC. You also notice that B is located halfway in between AC. AC/2 = B.

    Find the midpoint of the points in cartesian coordinates: (2,4) and (-7, 8).

    The Midpoint of points (2,4) and (-7, 8) would be the point in cartesian coordinates- ((-7 + 2)/2 , (8 + 4)/2) = (-5/2 , 6)
    Last edited: Jul 18, 2004
  5. Jul 18, 2004 #4


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    For the sake of simplicity, let [itex]AB[/itex] be a line segment on a two dimensional coordinate grid. [itex]A = (x_{1}, y_{1})[/itex] and [itex]B = (x_{2}, y_{2})[/itex]. The length [itex]l[/itex] of line segment [itex]AB[/itex] = [itex]\sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2}[/itex] by the Pythagorean theorem. A midpoint, by definition, is of half the length of the line segment, so we will define [itex]AQ[/itex] by point [itex]A[/itex] given previously and point [itex]Q = (x_{3}, y_{3})[/itex] such that the length

    [tex]\sqrt{(x_{3} - x_{1})^2 + (y_{3} - y_{1})^2} = \sqrt{(x_{2} - x_{3})^2 + (y_{2} - y_{3})^2} = l/2[/tex]

    Pick any numbers to fill in these values, then use induction to finish this part of the proof.

    I think you adequately proved that the points are on the same line.
    Last edited: Jul 18, 2004
  6. Jul 18, 2004 #5
    What is the difference between the application of the Pythagorean theorem to this proof and using the midpoint formula? I thought about using the midpoint formula but then thought against it.

    You guys are just brilliant. Thanks.
  7. Jul 18, 2004 #6


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    Midpoint formula follows from the Pythagoream theorem. Either term would work.
  8. Jul 18, 2004 #7
    "By the "definition of between", a point Q is between points A and B if and only if each of the following conditions hold.
    1.) A, Q, B are collinear.
    2.) AQ + BQ = AB"

    Incorrect! You must also say that |AQ|=|BQ|. The statment you wrote have no gurantee to be Q the mid-point. Find out the reason.
  9. Jul 18, 2004 #8


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    She knows that. That part of the proof was only meant to show that Q is colinear with A and B as well as somewhere between A and B, not necessarily that it is the midpoint.
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