Q. 54.2( 162 Votes )

# In Question 4, po

Answer :

Method 1:
Let A and B be the line segment and points P and Q be two different midpoints of AB. So,

AP = PB

And,

AQ = QB

And,

PB + AP = AB (It coincides with line segment AB)

Similarly,

QB + AQ = AB

Now,

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

2AP = AB   .....(i)

Similarly,

2AQ = AB .....(ii)

From (i) and (ii),

2AP = 2AQ (Since things which are equal to the same thing are equal to one another)

And as we know:

Things which are double of the same thing are equal to one another.

Therefore,

AP = AQ

Thus, P and Q are the same points.

This contradicts the fact that P and Q are two different midpoints AB.

Thus, it is proved that every line segment has one and only one midpoint.

Method 2: From the Figure,

AP + PB = AB.......................eq(i)

AQ + QB = AB.....................eq(ii)

From eq(i) and eq(ii)

AP + PB = AQ + QB

Now let AP = PB, and AQ = QB (as they are the midpoints, then)

2 AP = 2 AQ

AP = AQ

P = Q

Hence, there is only one midpoint.

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