Question

in the areas of two similar triangles are equal prove that are congruent

Answer 1

take triangle 1 as ABC and another as DEF
as area of triangle is 1/2×b×h
if the area is equal than breadth and height will equal of both triangle
by using Pythagoras theorem you can find the third side
then both third side of will be equal as b and h is equal
and hence both triangle I.e. ABC~=DEF
if you like please mark it as brainliest

Answer 2

mmiikk

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.