Proving AI = LI using Midpoint Postulate and Betweenness of Lines Theorem

  • Thread starter Thread starter OrbitalPower
  • Start date Start date
  • Tags Tags
    Geometric Proof
AI Thread Summary
The discussion focuses on proving that AI equals LI using the midpoint postulate and the betweenness of lines theorem. The proof begins with the midpoint I of segments AE and LC, establishing that AI equals IE and LI equals IC. Given AE equals LC, the equations AI + IE = AE and LI + IC = LC are utilized to derive relationships between the segments. By substituting the equalities, it is shown that 2AI equals 2LI, leading to the conclusion that AI equals LI. The proof effectively demonstrates the equality of the segments based on the geometric properties provided.
OrbitalPower
I have a simple geometric proof (first proofs) I can't finish. Looks like this:

A________L


C________E

suppose there's a straight line from l to c and a to e (to make an x) and a midpoint I.

It says: Given I is the midpoint of both \overline{AE} and \overline{LC}; AE = LC
Prove AI = LI

??

I'm not sure where to go except to start with the given. How do I complete (or start) the proof using the midpoint postulate and betweenness of lines theorm.
 
Physics news on Phys.org
Now, you DO know that I is the MIDPOINT of each line segment.

Therefore, you know that:
AI=IE and LI=IC

Furthermore,

you have the equations:
AE=LC (given)
AI+IE=AE (betweenness of points)
LI+IC=LC (the same)

Now, can you jumble about these 5 equations to get your result?
 
AE = LC (given)
AI+ IE = AE (betweeness of points)
LI + IC = LC (the same)
AI=IE
LI=IC
Therefore AI = LI because AE = LC so the segments would be equal.

That's what i could come up with with those equestions.
 
At this level, your logic should be a bit more thorough than that.
Since IE=AI and IC=LI, we have:

2AI=AE and 2LI=LC
Thus, we get:
2AI=2LI, implying AI=LI
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top