### Theory:

In earlier classes, we learned the construction of perpendicular bisector of a line segment, construction of an angle and construction of a bisector of a specified angle, all without including any reason. In this class, you will construct some of these and reasoning for why they are true.

Important!
Construct the bisector of an angle of $$60^\circ$$ and justify the construction.
Step 1: Draw $$\angle ABC = 60^\circ$$ using a protractor. Step 2: Taking $$B$$ as the centre and any convenient radius, draw an arc to intersect the rays $${BC}$$ and $${BA}$$ at $$D$$ and $$E$$, respectively. Step 3: Taking $$D$$ and $$E$$ as centres and the radius more than half of $$ED$$, draw arcs to intersect each other, say $$F$$. Step 4: Join $$BF$$, which is the bisector of $$\angle ABC = 60^\circ$$. JUSTIFICATION:

Join $$EF$$ and $$DF$$. In triangles $$BEF$$ and $$BDF$$,

$$BE = BD$$ [Radii of the same arc]

$$EF = DF$$ [Arcs of equal radii]

$$BF = BF$$ [Common side]

Therefore, $$\Delta BEF \cong \Delta BDF$$ [by $$SSS$$ congruence rule]

Corresponding parts of congruence triangles are congruent.

$$\Rightarrow \angle EBF = \angle DBF$$ - - - - - (I)

By construction, $$\angle ABC = \angle EBF + \angle DBF$$

$$\angle ABC = \angle EBF + \angle EBF$$ [Using equation (I)]

$$\angle ABC = 2 \angle EBF$$

$$\frac{1}{2} \angle ABC = \angle EBF$$

We know that $$\angle ABC = 60^\circ$$.

$$\frac{1}{2} \times 60^\circ = \angle EBF$$

$$30^\circ = \angle EBF$$

$$\Rightarrow \angle EBF = DBF = 30^\circ$$

Thus, $$BF$$ is a bisector of angle $$\angle ABC = 60^\circ$$.