if the areas of two similar triangles are equal then prove that the triangles are congruent
Answers
Answer:
→ ∆ABC ~ ∆DEF such that ar(∆ABC) = ar( ∆DEF) .
➡ To prove :-
→ ∆ABC ≅ ∆DEF .
➡ Proof :-
→ ∆ABC ~ ∆DEF . ( Given ) .
\begin{gathered}\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{gathered}
⟹
ar(△DEF)
ar(△ABC)
=
DE
2
AB
2
=
DF
2
AC
2
=
EF
2
BC
2
.........(1).
Now, ar(∆ABC) = ar( ∆DEF ) [ given ] .
\begin{gathered}\sf \implies \frac{ar( \triangle ABC )}{ ar( \triangle D EF )} = 1..........(2). \\\end{gathered}
⟹
ar(△DEF)
ar(△ABC)
=1..........(2).
▶ From equation (1) and (2), we get
\begin{gathered}\sf \implies \frac{AB^2}{DE^2} = \frac{AC^2}{DF^2} = \frac{BC^2}{EF^2} = 1 . \\\end{gathered}
⟹
DE
2
AB
2
=
DF
2
AC
2
=
EF
2
BC
2
=1.
⇒ AB² = DE² , AC² = DF² , and BC² = EF² .
[ Taking square root both sides, we get ] .
⇒ AB = DE , AC = DF and BC = EF .
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