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

(N/A) Let us assume two similar triangles as $\triangle ABC \sim \triangle PQR$.
According to the theorem of areas of similar triangles:
$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{AC}{PR}\right)^2$ $...(1)$
Given that,$\text{ar}(\triangle ABC) = \text{ar}(\triangle PQR)$.
Therefore,$\frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle PQR)} = 1$.
Substituting this value in equation $(1)$,we get:
$1 = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{AC}{PR}\right)^2$.
This implies:
$\left(\frac{AB}{PQ}\right)^2 = 1 \Rightarrow AB^2 = PQ^2 \Rightarrow AB = PQ$
$\left(\frac{BC}{QR}\right)^2 = 1 \Rightarrow BC^2 = QR^2 \Rightarrow BC = QR$
$\left(\frac{AC}{PR}\right)^2 = 1 \Rightarrow AC^2 = PR^2 \Rightarrow AC = PR$
Since all three corresponding sides are equal,by the $SSS$ (Side-Side-Side) congruence criterion,$\triangle ABC \cong \triangle PQR$.