Question

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

Answer 1

Let$△$ABC$\backsim$$△$ DEFand Area$△$ABC = Area$△$DEF)

Since the ratio of the area of two similair triangles is equal to the ratio of the squares on thier corresponding sides, we have.

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

$\Rightarrow$$\frac{A{B}^2}{D{E}^2}$ =$\frac{A{C}^2}{D{F}^2}$ =$\frac{B{C}^2}{E{F}^2}$ =1 [ Area($△$ABC = Area($△$DEF)]

$\Rightarrow$$A{B}^2$ =$D{E}^2$,$A{C}^2$=$D{F}^2$and $B{C}^2$ =$E{F}^2$

AB = DE,AC = DFand BC = EF

$△$ABC$\cong$ $△$DEF [By SSS congruence]