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# 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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