Question

If area of $△ABC=$ area of $△DEF$ then prove that $△ABC\cong △DEF$

Answer 1

Given $△ABC\sim △DEF$

We know that if two triangles are similar, then the ratio of there are equal to the square of the ratio of its corresponding sides.

$\frac{area△ABC}{area△DEF}={\left(\frac{BC}{EF}\right)}^2={\left(\frac{AB}{DE}\right)}^2={\left(\frac{AC}{DF}\right)}^2$

$\frac{area△DEF}{area△DEF}={\left(\frac{BC}{EF}\right)}^2={\left(\frac{AB}{DE}\right)}^2={\left(\frac{AC}{DF}\right)}^2$

since $area△ABC=area△DEF$

$\Rightarrow 1={\left(\frac{BC}{EF}\right)}^2={\left(\frac{AB}{DE}\right)}^2={\left(\frac{AC}{DF}\right)}^2$

$\Rightarrow 1={\left(\frac{BC}{EF}\right)}^2,1={\left(\frac{AB}{DE}\right)}^2,1{\left(\frac{AC}{DF}\right)}^2$

$\Rightarrow \frac{BC}{EF}=1,\frac{AB}{DE}=1,\frac{AC}{DF}=1$

$\Rightarrow BC=EF,AB=DE,AC=DF$

Hence by SSS congruency,

$△ABC\cong △DEF$

Hence proved.