If the areas of two similar triangles are equal, then the triangles are congruent, i.e. equal and similar triangles are congruent.

Solution :

Given : \(\triangle\) ABC ~ \(\triangle\) DEF such that \(area(\triangle ABC)\) = \(area(\triangle DEF)\)triangles

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

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