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 \(PQ = 13 \ cm\).
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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 \(A\).
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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
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.