Given:{tex}∆ABC\\sim∆PQR{/tex}and\xa0{tex}ar∆ABC=ar∆PQR{/tex}To prove:\xa0{tex}∆ABC\\cong∆PQR{/tex}Proof:\xa0{tex}∆ABC\\sim∆PQR{/tex}\xa0Also\xa0{tex}\\operatorname { ar } ( \\Delta A B C ) = \\operatorname { ar } ( \\Delta P Q R ){/tex}\xa0(given)or,\xa0{tex}\\frac { \\operatorname { ar } ( \\Delta A B C ) } { \\operatorname { ar } ( \\Delta P Q R ) } = 1{/tex}Or\xa0{tex}\\frac{AB^2}{PQ^2}=\\frac{BC^2}{QR^2}=\\frac{CA^2}{RP^2}=1{/tex}Or\xa0{tex}\\frac{AB}{PQ}=\\frac{BC}{QR}=\\frac{CA}{RP}=1{/tex}Hence we get thatAB=PQ,BC=QR and\xa0CA=RPHence\xa0{tex}∆ABC\\cong ∆PQR{/tex}