CASE II: \(AC - AB\), where \(AC > AB\)
Construct a triangle \(ABC\) in which \(BC = 6 \ cm\), \(\angle B = 45^\circ\) and \(AC - AB = 2 \ cm\).
Step 1: Draw a line segment \(BC = 6 \ cm\).
Pic 21.png
Step 2: Make \(\angle XBC = 45^\circ\).
Pic 22.png
Step 3: The line \(BX\) extended to opposite side of the line segment \(BC\). With \(B\) as the centre and \(2 \ cm\) as radius, draw an arc that cuts the extended \(BX\) at \(D\). Then, join \(CD\).
Pic 56.png
Step 4: Now, draw a perpendicular bisector \(EF\) of line segment \(CD\). Mark a point \(A\), where the perpendicular bisector intersects the ray \(BX\).
Pic 57.png
Step 5: Join \(AC\). Thus, \(ABC\) is the required triangle.
Pic 58.png
The point where \(EF\) cuts \(CD\) is marked as \(G\).
Pic 59.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, \(BD = AD - AB\) [by construction]
\(BD = AC - AB\) [using equation (I)]
Hence, the construction is justified.