Math, asked by 21122001, 1 year ago

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

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Answered by dragomegaman
3
Here is the required answer
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Answered by Anonymous
0

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.

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