We are given that:

$AB + BC = AC$

We need to show that $AB = BC$.

Proof:

Consider three points $A$, $B$, and $C$ on a straight line in that order. Since $AB + BC = AC$, the points must be collinear, meaning they lie on a straight line.

The condition $AB + BC = AC$ simply expresses that the total distance from $A$ to $C$ is the sum of the distances from $A$ to $B$ and from $B$ to $C$.

Now, let's assume that $B$ is the midpoint of segment $AC$. If $B$ is the midpoint, then the distances from $A$ to $B$ and from $B$ to $C$ must be equal. This is because the midpoint divides a segment into two equal parts.

Therefore:

$AB = BC$

Thus, $AB + BC = AC$ implies that $AB = BC$ if and only if $B$ is the midpoint of $AC$.

Conclusion:

Given the condition $AB + BC = AC$, if $B$ is the midpoint of $AC$, then $AB = BC$.

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Would you like further details or have any questions?

Here are some related questions:

1. What are other conditions for proving two line segments are equal?
2. How does midpoint definition relate to congruent segments?
3. Can we generalize this proof to vectors instead of line segments?
4. How can we express this proof using coordinate geometry?
5. What happens if $B$ is not the midpoint of $AC$?

Tip: In geometry, collinearity often simplifies many proofs related to distances and midpoints.