Prove that the ratio of the …

Given {tex}\triangle A B C{/tex} and {tex}\triangle PQR{/tex} in 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 {tex}\frac { a } { p } = \frac { b } { q } = \frac { c } { r }{/tex}
{tex}= \frac { a + b + c } { p + q + r }{/tex}
Proof Since {tex}\triangle ABC{/tex} and {tex}\triangle PQR{/tex} are similar, therefore their corresponding sides are proportional.
{tex}\therefore \quad \frac { a } { p } = \frac { b } { q } = \frac { c } { r } = k{/tex} (say) ...(i)
{tex}\Rightarrow \quad a = k p , b = k q{/tex} and {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} ...(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} [each equal to k].