Construct a triangle $ABC$ in which $BC = 8\,cm, \angle B = 45^{\circ}$ and $AB - AC = 3.5\,cm$.

(N/A) Steps of construction:
$I.$ Draw a ray $BX$.
$II.$ From $\overrightarrow{BX},$ cut off $\overline{BC} = 8\,cm$.
$III.$ Construct $\angle CBY = 45^{\circ}$.
$IV.$ From $\overrightarrow{BY},$ cut off $\overline{BD} = 3.5\,cm$.
$V.$ Join $D$ and $C$.
$VI.$ Draw the perpendicular bisector $PQ$ of $\overline{DC},$ which intersects $\overrightarrow{BY}$ at $A$.
$VII.$ Join $AC$.
Thus,$ABC$ is the required triangle.