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.
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)
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.