If a point \(C\) lies between two points \(A\) and \(B\) such that \(AC = BC\), then prove that \(AC = \frac{1}{2} AB\). Here, point \(C\) is called a mid-point of line segment \(AB\). Prove that every line segment has one and only one mid-point.
Proof:
Let there be two mid-points, \(C\) and \(D\).
\(C\) is the mid-point of \(AB\).
\(AC =\)
\(AC+AC=\) \(+AC\) (Equals are added on both sides) … (1)
Here, \((+ AC)\) coincides with \(AB\).
It is known that things which coincide with one another are equal to one another.
Therefore, \(+ AC = AB\) … (2)
It is also known that things which are equal to the same thing are equal to one another.
Therefore, from equations (1) and (2), we obtain
\(AC +\) \(= AB\)
That is, \( = AB\) … (3)
Similarly, by taking \(D\) as the mid-point of \(AB\), it can be proved that \(2AD= \) … (4)
From equation (3) and (4), we obtain
\(2AC = 2AD\)
Therefore, \(AC = AD\)
This is possible only when point \(C\) and \(D\) are representing a single point.
Hence, our assumption is wrong and there can be only one mid-point of a given line segment.