### 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 $$PQ = 13 \ cm$$.

Step 2: Construct $$\angle QPX = \angle B = 60^\circ$$, and $$\angle PQY = \angle C = 45^\circ$$.

Step 3: Now, draw bisectors of $$\angle QPX$$ and $$\angle PQY$$. Let these bisectors intersect at a point $$A$$.

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 $$C$$.

Step 5: Join $$AB$$ and $$AC$$. Thus, $$ABC$$ is the required triangle.

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 $$\angle PQY = \angle C$$.

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)

$$AB + CA = PB + CQ$$

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.