prove that every line has one and only one mid point
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Answered by
4
Let us consider, a line segment AB.
Assume that it has two midpoints say C and D
Recall that the midpoint of a line segment divides it into two equal parts
That is AC = BC and AD = DB
Since C is midpoint of AB, we have A, C and B are collinear
∴ AC + BC = AB → (1)
Similarly, we get AD + DB = AB → (2)
From (1) and (2), we get
AC + BC = AD + DB
2 AC = 2AD
∴ AC = AD
This is a contradiction unless C and D coincide.
Therefore our assumption that a line segment AB has two midpoints is incorrect.
Thus every line segment has one and only one midpoint.
hope it helped you...
Assume that it has two midpoints say C and D
Recall that the midpoint of a line segment divides it into two equal parts
That is AC = BC and AD = DB
Since C is midpoint of AB, we have A, C and B are collinear
∴ AC + BC = AB → (1)
Similarly, we get AD + DB = AB → (2)
From (1) and (2), we get
AC + BC = AD + DB
2 AC = 2AD
∴ AC = AD
This is a contradiction unless C and D coincide.
Therefore our assumption that a line segment AB has two midpoints is incorrect.
Thus every line segment has one and only one midpoint.
hope it helped you...
Abhi6611:
thanks behana ^-^ ^-^
Answered by
2
Let us consider a line segment AB.
Assume tht it has two midpoints C and D.
so,
AC=BC and AD=BD.
Since C is the midpoint A,C and B are collinear.
AC+BC=AB -(1) equation
AD+BD=AB -(2)equation
from (1) and (2) we get
AC+BC=AD+BD
2AC=2AD
ie AC=AD
This means C and D coincides.
Therefore every line segments have one and only one midpoint.
Assume tht it has two midpoints C and D.
so,
AC=BC and AD=BD.
Since C is the midpoint A,C and B are collinear.
AC+BC=AB -(1) equation
AD+BD=AB -(2)equation
from (1) and (2) we get
AC+BC=AD+BD
2AC=2AD
ie AC=AD
This means C and D coincides.
Therefore every line segments have one and only one midpoint.
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