Q5:

In Question 4, point $C$ is called a mid-point of line segment $AB$. Prove that every line segment has one and only one mid-point.

Step 1: Initial Setup & The Definition of a Mid-point

Consider a line segment $AB$. By geometric definition, a point $C$ is considered the mid-point of the line segment $AB$ if it lies on the segment between $A$ and $B$, and divides it into two equal parts. Mathematically, this is expressed as:

$AC = BC$

Since point $C$ lies on the line segment $AB$, the sum of the parts equals the whole [Per Euclid's Axiom 4: Things which coincide with one another are equal to one another, and Axiom 5: The whole is greater than the part]. Therefore:

$AC + BC = AB$

Substituting $BC$ with $AC$ yields:

$AC + AC = AB \implies 2AC = AB \implies AC = \frac{1}{2}AB \quad \text{--- (Equation 1)}$

Step 2: Formulating the Hypothesis for Proof by Contradiction

To prove that a line segment has one and only one mid-point, we will utilize a proof by contradiction. We begin by assuming the opposite of our desired conclusion.

Assumption: Let us assume that the line segment $AB$ has two distinct mid-points, namely $C$ and $D$.

Step 3: Applying Euclidean Axioms to the Assumption

If $D$ is also a mid-point of the line segment $AB$, it must satisfy the exact same geometric conditions as point $C$. Therefore, $D$ divides $AB$ into two equal parts ($AD = DB$). Following the identical algebraic derivation from Step 1, we obtain:

$AD = \frac{1}{2}AB \quad \text{--- (Equation 2)}$

Now, we analyze Equation 1 and Equation 2:

• From Equation 1: $AC = \frac{1}{2}AB$
• From Equation 2: $AD = \frac{1}{2}AB$

We apply Euclid’s First Axiom: "Things which are equal to the same thing are equal to one another."

Since both $AC$ and $AD$ are equal to $\frac{1}{2}AB$, they must be equal to each other:

$AC = AD$

Step 4: Geometric Interpretation and Resolution of the Contradiction

We have established that the distance from $A$ to $C$ is exactly equal to the distance from $A$ to $D$ along the same line segment $AB$.

Because points $C$ and $D$ both lie on the line segment $AB$ and are measured from the same origin point $A$ in the same direction, the equation $AC = AD$ implies that there is zero distance between point $C$ and point $D$.

By Euclid’s Fourth Axiom: "Things which coincide with one another are equal to one another." Conversely, since the lengths are equal and share the same initial point and direction, the endpoints must coincide. Therefore, point $C$ and point $D$ are not distinct; they are the exact same point.

This directly contradicts our initial assumption that $C$ and $D$ are two distinct mid-points. Because the assumption leads to a logical impossibility, the assumption must be false.

Final Solution: By proving that any assumed second mid-point must perfectly coincide with the first, we have rigorously demonstrated that every line segment possesses one, and only one, mid-point.