Let, AB be the line segment

Assume that points P and Q are the two different mid points of AB.

∴ P and Q are midpoints of AB.

Therefore,

AP = PB and AQ = QB.

PB+AP = AB (as it coincides with line segment AB)

Similarly, QB+AQ = AB.

Adding AP to the L.H.S and R.H.S of the equation AP = PB

AP + AP = PB + AP (If equals are added to equals, the wholes are equal.)

⇒ 2AP = AB — (i)

Similarly,

2 AQ = AB — (ii)

From (i) and (ii), Since R.H.S are same, we equate the L.H.S

2 AP = 2 AQ (Things which are equal to the same thing are equal to one another.)

⇒ AP = AQ (Things which are double of the same things are equal to one another.)

Thus, we conclude that P and Q are the same points.

This contradicts our assumption that P and Q are two different mid points of AB.

Thus, it is proved that every line segment has one and only one mid-point.

Hence Proved.