
#1
Jul1804, 01:03 AM

P: 566

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: 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
Jul1804, 01:08 AM

P: 566

Here is the proof my math tutor wrote:




#3
Jul1804, 01:20 AM

P: 100

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 ABC. 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) 



#4
Jul1804, 01:25 AM

Emeritus
PF Gold
P: 3,634

Midpoint proof
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. 



#5
Jul1804, 01:34 AM

P: 566

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. 



#6
Jul1804, 01:37 AM

Emeritus
PF Gold
P: 3,634

Midpoint formula follows from the Pythagoream theorem. Either term would work.




#7
Jul1804, 04:17 AM

P: 28

"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 midpoint. Find out the reason. 


Register to reply 
Related Discussions  
Midpoint rule  Calculus & Beyond Homework  3  
f preserves midpoint?  Calculus & Beyond Homework  5  
Derinving midpoint  General Math  2  
angular velocity of midpoint  Introductory Physics Homework  13  
What are the coordinates of point B?  General Math  3 