### Theory:

Important!
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 $$8 \ cm$$.

Step 2: Taking $$P$$ and $$Q$$ as centres and radius more than half of $$PQ$$, draw intersecting arcs above the line segment $$PQ$$, say $$R$$.

Step 3: Again, taking $$P$$ and $$Q$$ as centres and the same radius, draw intersecting arcs below the line segment $$PQ$$, say $$S$$.

Step 4: Join $$RS$$, which cuts $$PQ$$ at $$M$$. The line segment $$RS$$ is the perpendicular bisector of line segment $$PQ$$.

JUSTIFICATION:

Join $$PR$$, $$RQ$$, $$QS$$ and $$PS$$.

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 $$PQ$$.