Question

If areas of two similar triangles are equal, prove that they are congruent.

Answer 1

Applying the theorem of area of similar triangles to prove they are congruent:

Given that $\mathrm{area}\mathrm{of}∆ABC=\mathrm{area}\mathrm{of}∆DEF$

Then we can write it as,

$\frac{\mathrm{area}\mathrm{of}(∆ABC)}{\mathrm{area}\mathrm{of}(∆DEF)}=1$
According to the theorem of area of similar triangles if two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

Applying this in $∆ABC$ and $∆DEF$ we get,

$\frac{A{B}^2}{D{E}^2}=\frac{B{C}^2}{E{F}^2}=\frac{C{A}^2}{F{D}^2}=1$

Hence, $$\begin{gathered} AB=DE \\ BC=EF \\ CA=FD \end{gathered}$$

So by $SSS$ similarity criterion, we conclude that $△ABC\cong △DEF$

Hence, it is proven that the areas of two similar triangles are equal, then they are congruent.