Question

Prove that, if the areas of two similar triangles are equal, then they are congruent.

Answer 1

Given that:

$△ABC\sim △DEF$ and $ar\left(△ABC\right)=ar\left(△DEF\right)$

To prove:

$△ABC\cong △DEF$

Solution:

$\frac{Areaof△ABC}{Areaof△DEF}=\frac{A{B}^2}{D{E}^2}=\frac{B{C}^2}{E{F}^2}=\frac{A{C}^2}{D{F}^2}$

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

$A{B}^2=D{E}^2$

$AB=DE$

Similarly,

$B{C}^2=E{F}^2$

$BC=EF$

$A{C}^2=D{F}^2$

$AC=DF$

Now, in $△ABC$ and $△DEF,$

$AB=DE$

$BC=EF$

$AC=DF$

$\therefore$ $△ABC\cong △DEF$ (By SSS congruency criteria)