If the areas of two similar triangles are equal,

Solution:

Let $\Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$ and

area $(\Delta \mathrm{ABC}) \quad=$ area $(\Delta \mathrm{PQR}) \quad($ Given $)$

i.e., $\frac{\operatorname{area}(\Delta A B C)}{\operatorname{area}(\Delta P O R)}=1$

$\Rightarrow \frac{A B^{2}}{P Q^{2}}=\frac{B C^{2}}{Q R^{2}}=\frac{C A^{2}}{P R^{2}}=1$

$\Rightarrow \mathrm{AB}=\mathrm{PQ}, \mathrm{BC}=\mathrm{QR}$ and $\mathrm{CA}=\mathrm{PR}$

$\Rightarrow \Delta \mathrm{ABC} \cong \Delta \mathrm{PQR}$