Question

If in Q.2, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point.

Answer 1

Answer:

Solution:

Let, AB be the line segment

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

Therefore,

AP=PB ………(1)

and AQ = QB …..(2)

Also,

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

Similarly, QB + AQ = AB.

Now,

Adding AP to the L.H.S and R.H.S of the equation (1)

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

⇒ 2 AP = AB — (3)

Similarly,

2 AQ = AB — (4)

From (3) and (4), 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 midpoints of AB.

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

Hence Proved.

Hope it will be helpful :)

Answer 2

Answer:

Let there be two mid-points, C and D.

C is the mid-point of AB.

AC = CB

(Equals are added on both sides) …

AC + AC = BC + AC

(1)

Here, (BC + AC) coincides with AB. It is known that things which coincide with one another are equal to one another.

∴ BC + 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 + AC = AB

⇒ 2AC = AB … (3)

Similarly, by taking D as the mid-point of AB, it can be proved that

2AD = AB … (4)

From equation (3) and (4), we obtain

2AC = 2AD (Things which are equal to the same thing are equal to one another.)

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

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.