# 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 $$area(\triangle ABC)$$ = $$area(\triangle DEF)$$

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