Theory:

Construct a triangle \(ABC\) in which \(\angle B = 60^\circ\), \(\angle C = 45^\circ\) and \(AB + BC + CA = 13 \ cm\).
 
Step 1: Draw a line segment \(PQ = 13 \ cm\).
 
Pic 26.png
 
Step 2: Construct \(\angle QPX = \angle B = 60^\circ\), and \(\angle PQY = \angle C = 45^\circ\).
 
Pic 27.png
 
Step 3: Now, draw bisectors of \(\angle QPX\) and \(\angle PQY\). Let these bisectors intersect at a point \(A\).
 
Pic 28.png
 
Step 4: Construct perpendicular bisectors \(RS\) of line segment \(AP\) and \(TU\) of line segment \(AQ\). The points where \(RS\) intersects \(PQ\) as \(B\) and \(TS\) intersects \(AQ\) as \(C\).
 
Pic 29.png
 
Step 5: Join \(AB\) and \(AC\). Thus, \(ABC\) is the required triangle.
 
Pic 30.png
 
JUSTIFICATION:
 
The point at which \(RS\) cuts \(AP\) as \(M\) and \(TU\) cuts \(AQ\) as \(N\).
 
We need to prove \(AB + BC + CA = PQ\), \(\angle QPX = \angle B\) and \(\angle PQY = \angle C\).
 
Pic 30_A.png
 
In triangles \(BMP\) and \(BMA\):
 
\(PM = AM\) [Since \(RS\) is the bisector of \(AP\)]
 
\(\angle BMP = \angle BMA\) [Since \(RS\) is perpendicular to \(AP\)]
 
\(MB = MB\) [Common side]
 
Therefore, \(\Delta BMP \cong \Delta BMA\) [by \(SAS\) congruence rule]
 
\(\Rightarrow PB = AB\) [by CPCT] - - - - - (I)
 
Similarly, for triangles \(CNA\) and \(CNQ\), \(CQ = CA\) - - - - - (II)
 
Add equation (I) and (II).
 
\(AB + CA = PB + CQ\)
 
Add \(BC\) on both sides.
 
\(AB + BC + CA = PB + BC + CQ\)
 
\(AB + BC + CA = PQ\)
 
In triangle \(ABP\), \(PB = AB\)
 
Angles opposite to equal sides are equal.
 
\(\Rightarrow \angle APB = \angle BAP\)
 
The exterior angle is equal to the sum of opposite interior angles.
 
\(\Rightarrow \angle ABC = \angle APB + \angle BAP\)
 
\(\Rightarrow \angle ABC = \angle APB + \angle APB\)
 
\(\Rightarrow \angle ABC = 2 \angle APB\)
 
\(\Rightarrow \angle ABC = \angle XPB = \angle XPQ\) [Since \(AP\) is an angle bisector of \(\angle QPX\)]
 
Similarly, \(\angle ACB = \angle PQY\).
 
Hence, we justified.