Solution :
Given : \(\triangle\) ABC ~ \(\triangle\) DEF such that \(area(\triangle ABC)\) = \(area(\triangle DEF)\)
To Prove : \(\triangle\) ABC \(\cong\) \(\triangle\) DEF
Proof : Since, the ratio of the area of two similar triangles is equal to the ratio of the square of two corresponding sides.
\(area(\triangle ABC)\over area(\triangle DEF)\) = \({AB}^2\over {DE}^2\) = \({AC}^2\over {DF}^2\) = \({BC}^2\over {EF}^2\)
Given, \(area(\triangle ABC)\) = \(area(\triangle DEF)\)
\({AB}^2\over {DE}^2\) = \({AC}^2\over {DF}^2\) = \({BC}^2\over {EF}^2\) = 1
\({AB}^2\) = \({DE}^2\), \({AC}^2\) = \({DF}^2\) and \({BC}^2\) = \({EF}^2\)
AB = DE, AC = DF and BC = EF
By SSS theorem of congruence,
\(\triangle\) ABC \(\cong\) \(\triangle\) DEF
