Theory:

There are two cases while constructing a triangle using its base, a base angle and the difference of other two sides.

Suppose you are constructing triangle $$ABC$$, for a given base $$BC$$, a base angle $$\angle B$$, then there are two possibilities for a difference of the other two sides. One is $$AB - AC$$, and the other is $$AC - AB$$. Let us discuss the first case in this material.
CASE I: $$AB - AC$$, where $$AB > AC$$
Construct a triangle $$ABC$$ in which $$BC = 6 \ cm$$, $$\angle B = 45^\circ$$ and $$AB - AC = 2 \ cm$$.

Step 1: Draw a line segment $$BC = 6 \ cm$$.

Step 2: Make $$\angle XBC = 45^\circ$$.

Step 3: Given that, $$AB - AC = 2 \ cm$$. Let's take $$2 \ cm$$ 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 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, $$BD = AB - AD$$ [by construction]

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

Hence, the construction is justified.