Prove that the ratio of the perimeter of two similar triangles is equal to the corresponding sides

Prove that the ratio of the perimeter of two similar triangles is equa

1.	Prove that the ratio of the perimeter of two similar triangles is equal to the corresponding sides
Answer» Given\xa0{tex}\\triangle A B C{/tex}\xa0and\xa0{tex}\\triangle PQR{/tex}\xa0in Which BC= a, CA = b, AB = c and QR = p, RP = q, PQ = r.Also, {tex}\\triangle A B C \\sim \\triangle P Q R{/tex}To Prove\xa0{tex}\\frac { a } { p } = \\frac { b } { q } = \\frac { c } { r }{/tex}{tex}= \\frac { a + b + c } { p + q + r }{/tex}Proof Since\xa0{tex}\\triangle ABC{/tex}\xa0and\xa0{tex}\\triangle PQR{/tex}\xa0are similar, therefore their corresponding sides are proportional.{tex}\\therefore \\quad \\frac { a } { p } = \\frac { b } { q } = \\frac { c } { r } = k{/tex}\xa0(say) ...(i){tex}\\Rightarrow \\quad a = k p , b = k q{/tex}\xa0and\xa0{tex}c = k r{/tex}{tex}\\therefore \\quad \\frac { \\text { perimeter of } \\triangle A B C } { \\text { perimeter of } \\triangle P Q R }{/tex}{tex}= \\frac { a + b + c } { p + q + r }{/tex}{tex}= \\frac { k p + k q + k r } { p + q + r }{/tex}{tex}= \\frac { k ( p + q + r ) } { ( p + q + r ) } = k{/tex}\xa0...(ii)From (i) and (ii), we get{tex}\\frac { a } { p } = \\frac { b } { q } = \\frac { c } { r }{/tex}{tex}= \\frac { a + b + c } { p + q + r }{/tex}{tex}= \\frac { \\text { perimeter of } \\triangle A B C } { \\text { perimeter of } \\triangle P Q R }{/tex}\xa0[each equal to k].