Question

if the area of two similar triangle are equal then prove that they are congruent

Answer 1

Use the theorem that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides , then prove that they are congruent.
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Solution:

[Fig is in the attachment]

Given: ΔABC ~ ΔPQR. &

ar ΔABC =ar ΔPQR

To Prove: ΔABC ≅ ΔPQR

Proof: Since, ΔABC ~ ΔPQR
ar ΔABC =ar ΔPQR. (given)

ΔABC / ar ΔPQR = 1

⇒ AB²/PQ² = BC²/QR² = CA²/PR² = 1

[ USING THEOREM OF AREA OF SIMILAR TRIANGLES]

⇒ AB= PQ , BC= QR & CA= PR

Thus, ΔABC ≅ ΔPQR
[BY SSS criterion of congruence]

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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.