Theory:

Construct a triangle $$ABC$$ in which $$BC = 8 \ cm$$, $$\angle B = 60^\circ$$ and $$AB + AC = 12 \ cm$$.

Step 1: Draw a line segment $$BC = 8 \ cm$$.

Step 2: Make $$\angle XBC = 60^\circ$$.

Step 3: Given that, $$AB + AC = 12 \ cm$$. Let's take $$12 \ cm$$ as the radius, with $$B$$ as centre, draw an arc that cuts $$BX$$ at $$D$$. Then, join $$CD$$.

Step 4: Now, draw a perpendicular bisector $$EF$$ of line segment $$CD$$. Mark a point $$A$$, where perpendicular bisector intersects $$BD$$.

Step 5: Join $$AC$$. Thus, $$ABC$$ is the required triangle.

JUSTIFICATION:

The point where $$EF$$ cuts $$CD$$ is marked as $$G$$.

In triangles $$AGD$$ and $$AGC$$:

$$DG = GC$$ [Since $$EF$$ bisects $$CD$$ by construction]

$$\angle AGD = \angle AGC$$ [Since $$EF$$ perpendicular $$CD$$ by construction]

$$AG = AG$$ [Common side]

Therefore, $$\Delta AGD \cong \Delta AGC$$ [by $$SAS$$ congruence rule]

$$\Rightarrow AD = AC$$ [by CPCT] - - - - - - (I)

Now, $$AB = BD - AD$$

$$AB = BD - AC$$ [Using (I)]

$$AB + AC = BD = 12 \ cm$$

Therefore, our construction is justified.