Theory:

Important!
To know how to construct a perpendicular bisector, click here.
Construct a perpendicular bisector of a line segment \(PQ\) of length \(8 \ cm\) and justify the construction.
Step 1: Draw a line segment \(PQ\) of length \(8 \ cm\).
 
Pic6.PNG
 
Step 2: Taking \(P\) and \(Q\) as centres and radius more than half of \(PQ\), draw intersecting arcs above the line segment \(PQ\), say \(R\).
 
Pic7.PNG
 
Step 3: Again, taking \(P\) and \(Q\) as centres and the same radius, draw intersecting arcs below the line segment \(PQ\), say \(S\).
 
Pic8.PNG
 
Step 4: Join \(RS\), which cuts \(PQ\) at \(M\). The line segment \(RS\) is the perpendicular bisector of line segment \(PQ\).
 
Pic9.PNG
 
JUSTIFICATION:
 
Join \(PR\), \(RQ\), \(QS\) and \(PS\).
 
Pic10.PNG
 
In triangles \(RPS\) and \(RQS\):
 
\(RP = RQ\) [Arcs of equal radii]
 
\(PS = SQ\) [Arcs of equal radii]
 
\(RS = RS\) [Common side]
 
Therefore, \(\Delta RPS \cong \Delta RQS\) [by \(SSS\) congruence rule]
 
Corresponding parts of congruence triangles are congruent.
 
\(\Rightarrow \angle PRM = \angle QRM\) - - - - - - - (I)
 
Now, in triangles \(PRM\) and \(QRM\):
 
\(PR = RQ\) [Arcs of equal radii]
 
\(\angle PRM = \angle QRM\) [Using equation (I)]
 
\(RM = RM\) [Common side]
 
Therefore, \(\Delta PRM \cong \Delta QRM\) [by \(SAS\) congruence rule]
 
Corresponding parts of congruence triangles are congruent.
 
\(\Rightarrow PM = MQ\) and \(\angle PMR = \angle QMR\) - - - - (II)
 
The sum of the adjacent angles in a straight line is \(180^\circ\).
 
\(\Rightarrow \angle PMR + \angle QMR = 180^\circ\)
 
\(\Rightarrow \angle PMR + \angle PMR = 180^\circ\) [Using equation (II)]
 
\(\Rightarrow 2 \angle PMR = 180^\circ\)
 
\(\Rightarrow \angle PMR = 90^\circ\)
 
So, \(\Rightarrow \angle PMR = \angle QMR = 90^\circ\).
 
We get \(PM = MQ\) and \(\angle PMR = \angle QMR = 90^\circ\).
 
Therefore, \(RS\) is the perpendicular bisector of \(PQ\).