Proving the Existence of a Midpoint on Every Segment | Segment Midpoint Theorem

[Bashyboy]

Homework Statement

Prove that every segment has a midpoint.

Homework Equations

The Attempt at a Solution

I first began with some arbitrary segment $AB$ in the plane, and then constructed the line $\overset{\leftrightarrow}{AB} = \ell$ from these two points. I then used the theorem which states that, given two points $A$ and $B$ on a line, there exists a point $C$ not on the line such that $\triangle ACB$ is an equilateral triangle. I then was going to use the angle bisector theorem to form a bisector which would intersect the segment $AB$ at a point $D$. Using the side-side-side criterion of a triangle, I could conclude that $AD \cong DB$.

However, there are few issues with this. Firstly, how do I know the angle bisector will intersect the segment $AB$; why is it not possible that the angle bisector $\overrightarrow{CD}$ to curve and loop around in such a way that it never intersects the segment $AB$ nor the line $\ell$? Secondly, even if it does intersect $\ell$, how do I know that $D$ is between $A$ and $B$

For the second issue, I tried a proof by contradiction, but I couldn't identify any contradiction.

 

[SteamKing]

Homework Statement

Prove that every segment has a midpoint.

Homework Equations

The Attempt at a Solution

I first began with some arbitrary segment $AB$ in the plane, and then constructed the line $\overset{\leftrightarrow}{AB} = \ell$ from these two points. I then used the theorem which states that, given two points $A$ and $B$ on a line, there exists a point $C$ not on the line such that $\triangle ACB$ is an equilateral triangle. I then was going to use the angle bisector theorem to form a bisector which would intersect the segment $AB$ at a point $D$. Using the side-side-side criterion of a triangle, I could conclude that $AD \cong DB$.

However, there are few issues with this. Firstly, how do I know the angle bisector will intersect the segment $AB$; why is it not possible that the angle bisector $\overrightarrow{CD}$ to curve and loop around in such a way that it never intersects the segment $AB# nor the line$\ell$? Secondly, even if it does intersect$\ell$, how do I know that$D$is between$A$and$B##

The bisector of an angle is a ray which extends from the apex of the angle to infinity. By definition, a ray is a line, without any loops or whorls in it.

http://www.mathopenref.com/ray.html

 

[Ray Vickson]

Homework Statement

Prove that every segment has a midpoint.

Homework Equations

The Attempt at a Solution

I first began with some arbitrary segment $AB$ in the plane, and then constructed the line $\overset{\leftrightarrow}{AB} = \ell$ from these two points. I then used the theorem which states that, given two points $A$ and $B$ on a line, there exists a point $C$ not on the line such that $\triangle ACB$ is an equilateral triangle. I then was going to use the angle bisector theorem to form a bisector which would intersect the segment $AB$ at a point $D$. Using the side-side-side criterion of a triangle, I could conclude that $AD \cong DB$.

However, there are few issues with this. Firstly, how do I know the angle bisector will intersect the segment $AB$; why is it not possible that the angle bisector $\overrightarrow{CD}$ to curve and loop around in such a way that it never intersects the segment $AB$ nor the line $\ell$? Secondly, even if it does intersect $\ell$, how do I know that $D$ is between $A$ and $B$

For the second issue, I tried a proof by contradiction, but I couldn't identify any contradiction.View attachment 89701

What is wrong with the standard construction, as in http://www.mathopenref.com/constbisectline.html ?

 

[Bashyboy]

The bisector of an angle is a ray which extends from the apex of the angle to infinity. By definition, a ray is a line, without any loops or whorls in it.

http://www.mathopenref.com/ray.html

That seems to be a rather vague definition. The definition I am using is the following: Let $A$ and $B$ be two points. The ray starting at $A$ and going through $B$ is the set of all points $X$ such that either (1) $X$ is $A$ or $B$, (2) $X$ is between $A$ and $B$, or $B$ is between $A$ and $X$. Notice, such a definition does not mention infinity and does not preclude the possibility of loops and curves.

What is wrong with the standard construction, as in http://www.mathopenref.com/constbisectline.html ?

Well, it is proposed by our professor that we try to rely on recently proven theorems, such as being able to construct an equilateral triangle from three points, rather than always resorting to drawing circles.

 

[vela]

That seems to be a rather vague definition. The definition I am using is the following: Let $A$ and $B$ be two points. The ray starting at $A$ and going through $B$ is the set of all points $X$ such that either (1) $X$ is $A$ or $B$, (2) $X$ is between $A$ and $B$, or $B$ is between $A$ and $X$. Notice, such a definition does not mention infinity and does not preclude the possibility of loops and curves.

How do you get that it doesn't preclude the possibility of loops and curls? For instance, if the ray loops or curls between A and B, then there will be points on the curve that aren't between A and B. Between means a point is on the line segment connecting A and B, not that it's on any old curve that passes through A and B.