if the area of two similar triangles are equal then prove that its congruent?

$\mathrm{The} \mathrm{theorem} \mathrm{on} \mathrm{areas} \mathrm{of} \mathrm{similar} \mathrm{triangles} \mathrm{says}:$$\mathrm{The} \mathrm{ratio} \mathrm{of} \mathrm{the} \mathrm{areas} \mathrm{of} \mathrm{two} \mathrm{similar} \mathrm{triangles} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{square} \mathrm{of} \mathrm{the} \mathrm{ratio} \mathrm{of} \mathrm{their} \mathrm{corresponding} \mathrm{sides}.$$\mathrm{Consider} 2 \mathrm{similar} \mathrm{triangles} \mathrm{ABC} \mathrm{and} \mathrm{PQR},$$\frac{\mathrm{area} \mathrm{of} \mathrm{\Delta ABC}}{\mathrm{area} \mathrm{of} \mathrm{\Delta PQR}}={\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)}^2={\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)}^2={\left(\frac{\mathrm{CA}}{\mathrm{RP}}\right)}^2$$\mathrm{As} \mathrm{the} \mathrm{areas} \mathrm{are} \mathrm{equal}, {\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)}^2={\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)}^2={\left(\frac{\mathrm{CA}}{\mathrm{RP}}\right)}^2=1$$\mathrm{So} \mathrm{AB} = \mathrm{PQ}; \mathrm{BC}=\mathrm{QR};\mathrm{CA}=\mathrm{RP}$$\mathrm{therefore} \mathrm{triangles} \mathrm{are} \mathrm{congruent} \mathrm{by} \mathrm{SSS}.$">