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 \(BC = 8 \ cm\).
 
Pic16.PNG
 
Step 2: Make \(\angle XBC = 60^\circ\).
 
Pic17.PNG
 
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 \(CD\).
 
Pic18.PNG
 
Step 4: Now, draw a perpendicular bisector \(EF\) of line segment \(CD\). Mark a point \(A\), where perpendicular bisector intersects \(BD\).
 
Pic19.PNG
 
Step 5: Join \(AC\). Thus, \(ABC\) is the required triangle.
 
Pic20.png
 
JUSTIFICATION:
 
The point where \(EF\) cuts \(CD\) is marked as \(G\).
 
Pic60.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, \(AB = BD - AD\)
 
\(AB = BD - AC\) [Using (I)]
 
\(AB + AC = BD = 12 \ cm\)
 
Therefore, our construction is justified.