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 \(BC = 6 \ cm\).
 
Pic21.PNG
 
Step 2: Make \(\angle XBC = 45^\circ\).
 
Pic22.PNG
 
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 \(CD\).
 
Pic23.PNG
 
Step 4: Now, draw a perpendicular bisector \(EF\) of line segment \(CD\). Mark a point \(A\), where the perpendicular bisector intersects the ray \(BX\).
 
Pic24.PNG
 
Step 5: Join \(AC\). Thus, \(ABC\) is the required triangle.
 
Pic25.PNG
 
JUSTIFICATION:
 
The point where \(EF\) cuts \(CD\) is marked as \(G\).
 
Pic25_A.PNG
 
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.