Question

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

Answer 1

Given,

to prove that, if area of two similar triangles are equal then, they are congruent.

Let us consider, the triangles

as $△ABC$ and $△DEF$.

also given, both the triangles are similar i.e, $△ABC\sim △DEF$ and there areas are also equal it means $ar△ABC=ar△DEF$

Required to prove (R.T.P) :- Both the triangles $[△ABC\cong △DEF]$ are congruent.

Proof :- from the given $△ABC\sim △DEF$, as we already know that when two triangles are similar then, their ratios of areas are equal to ratio of squares of their corresponding sides. So, we get

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

$\frac{ar△DEF}{ar△DEF}={\left(\frac{BC}{EF}\right)}^2={\left(\frac{AB}{DE}\right)}^2={\left(\frac{AC}{DF}\right)}^2$ [ $\because$ Given, $ar△ABC=ar△DEF$]

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

Taking them separately we get,

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

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

$\iff BC=EF$ $\iff AB=DE$ $\iff AC=DF$

So, now let us consider both $△ABC$ and $△DEF$ we got $EF=BCAB=DE$ and , $AC=DF$

Hence It is by $SSS$ congruency

$\therefore △ABC\cong △DEF$

Hence proved