Question

If the area of two similar triangle are equal prove that they are congruent?

Answer 1

If the areas of two similar triangles are equal, prove that they are congruent. If two triangles are similar, then (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion)
then U can easily prove that
hop its helpful

Answer 2

Step-by-step explanation:

Given :-

→ ∆ABC ~ ∆DEF such that ar(∆ABC) = ar( ∆DEF) .

➡ To prove :-

→ ∆ABC ≅ ∆DEF .

➡ Proof :-

→ ∆ABC ~ ∆DEF . ( Given ) .

$\begin{lgathered}\tiny \sf \implies \frac{ar( \triangle ABC )}{ ar( \triangle D EF )} = \frac{AB^2}{DE^2} = \frac{AC^2}{DF^2} = \frac{BC^2}{EF^2} .........(1) . \\\end{lgathered}$

Now, ar(∆ABC) = ar( ∆DEF ) [ given ] .

$\begin{lgathered}\sf \implies \frac{ar( \triangle ABC )}{ ar( \triangle D EF )} = 1..........(2). \\\end{lgathered}$

▶ From equation (1) and (2), we get,,,

$\begin{lgathered}\sf \implies \frac{AB^2}{DE^2} = \frac{AC^2}{DF^2} = \frac{BC^2}{EF^2} = 1 . \\\end{lgathered}$

⇒ AB² = DE² , AC² = DF² , and BC² = EF² .

[ Taking square root both sides, we get ] .

⇒ AB = DE , AC = DF and BC = EF .

$\large\pink{ \boxed{ \tt \therefore \triangle ABC \cong \triangle D EF .}}$

[ by SSS-congruency ] .

Hence, it is proved.