Proof:

Given: In the figure, point C lies between points A and B.

To prove: AB > AC

Proof:

1. Construction:
- Draw a line segment AB with points A and B.
- Place point C between points A and B.

2. Euclid's First Postulate:
- According to Euclid's First Postulate, we can draw a line segment between any two points. Therefore, we have line segment AC and line segment CB.

3. Euclid's Second Postulate:
- Euclid's Second Postulate states that we can extend a line segment indefinitely in a straight line.
- Therefore, we can extend line segment AB to form line segment ACB.

4. Definition of a Line Segment:
- A line segment is the shortest distance between two points.
- Since points A and B are fixed, the line segment AB is of a fixed length.

5. Euclid's Fifth Postulate:
- Euclid's Fifth Postulate, also known as the Parallel Postulate, states that through a point not on a given line, there is exactly one line parallel to the given line.
- In this case, line segment ACB is not parallel to line segment AB, as they intersect at point C.

6. Triangle Inequality Theorem:
- According to the Triangle Inequality Theorem, for any triangle, the sum of the lengths of any two sides is greater than the length of the third side.
- Applying the Triangle Inequality Theorem to triangle ABC:
- AB + BC > AC
- Since AB is the fixed length of the line segment AB, we can rewrite the inequality as:
- AB + BC > AB
- Subtracting AB from both sides, we get:
- BC > 0

7. Conclusion:
- From step 6, we can conclude that BC is greater than 0, which is always true.
- Therefore, AB + BC > AC is always true.
- Simplifying the inequality, we have AB > AC.

Hence, we have proved that AB > AC using Euclid's principles and the Triangle Inequality Theorem.