In Question 4, point $ \mathrm{C} $ is called a mid-point of line segment $ \mathrm{AB} $. Prove that every line segment has one and only one mid-point.
Given:
Point $C$ is the midpoint of $\overline{AB}$.
To do:
We have to prove that every line segment has one and only one midpoint.
Solution:
Let us assume points $C$ and $D$ are two mid-points of $\overline{AB}$.
Since, $C$ and $D$ are midpoints of $\overline{AB}$.
We get,
$AC=CB$ and $AD=BD$
According to Euclid's Axiom
We get,
$AC+CB=AB$ (Since, $AC+CB$ coincides with $AB$)
Similarly, we get,
$AD+BD=AB$ (Since, $AD+BD$ coincides with $AB$)
Now,
By adding $AC$ on both sides of $AC=CB$
We get,
$AC+AC=CB+AC$ (Since, if equals are added to equals the wholes are equals.)
This implies,
$2AC=AB$...........(i)
In a similar way, we get,
$AD+AD=DB+AD$ (Since, if equals are added to equals the wholes are equals.)
This implies,
$2AD=AB$.............(ii)
From (i) and (ii)
We got R.H.S as the same
Therefore,
Let us equate L.H.S of (i) and (ii)
We get,
$2AC=2AD$ (According to Euclid's Axiom: Things which are equal to the same thing are equal to one another.)
Therefore,
$AC=AD$(According to Euclid's Axiom: Things which are double of the same things are equal to one another.)
Therefore,
We can say that Points $C$ and $D$ are the same points.
Therefore,
Our assumption that $C$ and $D$ are two different midpoints is False.
Hence, every line segment has only one midpoint.
Hence proved.
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