Question

If the areas of two similar triangles are equal, prove that they are congruent.

Answer 1

Hint: There is given for this question, if the areas of two similar triangles are equal. Hence, we are going to use two similar triangles and its properties to prove they are congruent.
SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent.

Complete step-by-step answer:
It is given that if the areas of two similar triangles are equal. Then we have to prove they are congruent.
Let, ABC & DEF are two similar triangles.
That is, \[\Delta ABC \sim \Delta DEF\]
Given that, the areas of the triangles ABC & DEF are equal.
That is,\[{\text{Area}}(\Delta ABC) = {\text{Area}}(\Delta DEF)\]

We need to prove that the triangles are congruent.
That is, \[\Delta ABC \cong \Delta DEF\] .
Let us consider the given \[\Delta ABC \sim \Delta DEF\]
We know that if two triangle are similar and area of the triangles are equal
Ratio of areas is equal to square of ratio of its corresponding sides
\[ \Rightarrow \dfrac{{{\text{Area}}(\Delta ABC)}}{{{\text{Area}}(\Delta DEF)}} = {\left( {\dfrac{{BC}}{{EF}}} \right)^2} = {\left( {\dfrac{{AB}}{{DE}}} \right)^2} = {\left( {\dfrac{{AC}}{{DF}}} \right)^2}\]
Since, it is given that \[{\text{Area}}(\Delta ABC) = {\text{Area}}(\Delta DEF)\]
\[ \Rightarrow 1 = {\left( {\dfrac{{BC}}{{EF}}} \right)^2} = {\left( {\dfrac{{AB}}{{DE}}} \right)^2} = {\left( {\dfrac{{AC}}{{DF}}} \right)^2}\]
We will equate each term to 1,
\[ \Rightarrow {\left( {\dfrac{{BC}}{{EF}}} \right)^2} = 1\]
Taking square root on both sides,
\[
   \Rightarrow \dfrac{{BC}}{{EF}} = 1 \\
   \Rightarrow BC = EF........(1) \\
 \]
\[ \Rightarrow {\left( {\dfrac{{AB}}{{DE}}} \right)^2} = 1\]
Taking square root on both sides,
\[
   \Rightarrow \dfrac{{AB}}{{DE}} = 1 \\
   \Rightarrow AB = DE......(2) \\
 \]
\[ \Rightarrow {\left( {\dfrac{{AC}}{{DF}}} \right)^2} = 1\]
Taking square root on both sides,
\[
   \Rightarrow \dfrac{{AC}}{{DF}} = 1 \\
   \Rightarrow AC = DF........(3) \\
 \]
Thus we get,
\[ \Rightarrow BC = EF\]
\[ \Rightarrow AB = DE\]
\[ \Rightarrow AC = DF\]
Thus, in the triangles, \[\Delta ABC\] and \[\Delta DEF\]
\[ \Rightarrow BC = EF\]
\[ \Rightarrow AB = DE\]
\[ \Rightarrow AC = DF\]
Hence, by SSS congruence
\[\therefore \Delta ABC \cong \Delta DEF\].
Thus we get, when the areas of two similar triangles are equal, they are congruent.

Note: If an angle of one triangle is congruent to the corresponding angle of another triangle and the lengths of the sides including these angles are in proportion, the triangles are similar. The corresponding sides of similar triangles are in proportion.
We have used the following theorem.
Theorem: If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.