### Theory:

CASE II: $$AC - AB$$, where $$AC > AB$$
Construct a triangle $$ABC$$ in which $$BC = 6 \ cm$$, $$\angle B = 45^\circ$$ and $$AC - AB = 2 \ cm$$.

Step 1: Draw a line segment $$BC = 6 \ cm$$. Step 2: Make $$\angle XBC = 45^\circ$$. Step 3: The line $$BX$$ extended to opposite side of the line segment $$BC$$. With $$B$$ as the centre and $$2 \ cm$$ as radius, draw an arc that cuts the extended $$BX$$ at $$D$$. Then, join $$CD$$. Step 4: Now, draw a perpendicular bisector $$EF$$ of line segment $$CD$$. Mark a point $$A$$, where the perpendicular bisector intersects the ray $$BX$$. 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, $$BD = AD - AB$$ [by construction]

$$BD = AC - AB$$ [using equation (I)]

Hence, the construction is justified.