Question

if he area of two similar triangles are equal prove hat they are congruent

Answer 1

$\bold{\huge{\underline{CORRECT \:QUESTION}}}$

if area of two similar triangles are equal prove hat they are congruent

___________________

$\bold{\huge{\underline{ANSWER:-}}}$

Consider on two triangles :-

ΔABC and ΔDEF

Again;

∆ABC~∆DEF

which are similar to each other.

Given:-Ar(ΔABC = Ar(ΔDEF)

$\bold{NOTE:-}$
That ratio of area of two similar triangle is equal to the ratio of squares of their corresponding sides.

$\bold{\implies{ \frac{ar(abc)}{ar(def)} ={ ( \frac{AB}{DE})}^{2}}} = { \bold{ ( \frac{BC}{EF})}^{2}} = { \bold{ ( \frac{AC}{DF})}^{2}}$

$\implies \bold{1 = ( \frac{AB}{DE})}^{ \cancel{2}} = { \bold{ ( \frac{BC}{EF})}^{ \cancel{2}}} = { \bold{ ( \frac{AC}{DF})}^{\cancel{2}}} \\ \\ \\ \\ \implies\bold{1 = \frac{AB}{de} = \frac{BC}{EF} = \frac{AC}{DF}} \\ \\ \\ \implies \bold{ \frac{AB}{DE} = 1 \: \: , \frac{BC}{EF} = 1 \: \: , \frac{AC}{DF} = 1} \\ \\ \\ \implies\bold{AB = DE \: , \: BC = EF , \: \: AC= DF} ,$

Here,All sides remaining same !!

$\bold{Hence, It is proved by below conclusion}$

∴ All the sides of ΔABC are equal to the corresponding sides of ΔDEF.

Using Side Side and Side; Congruent Property

$\bold{Finally Proved}$

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.