side AB and BC and median …

It is given that:
 {tex}\frac { A B } { P Q } = \frac { B C } { Q R } = \frac { A D } { P M }{/tex}
{tex}\Rightarrow \quad \frac { A B } { P Q } = \frac { A D } { P M }{/tex}{tex}= \frac { B C } { Q R } = \frac { \frac { 1 } { 2 } B C } { \frac { 1 } { 2 } Q R } = \frac { B D } { Q M }{/tex} ....................................(i)
In {tex}\triangle ABD{/tex} and {tex}\triangle PQM{/tex}, we have
{tex}\frac { A B } { P Q } = \frac { A D } { P M } = \frac { B D } { Q M }{/tex}           [from(i)]
{tex}\therefore \quad \triangle A B D \sim \triangle P Q M{/tex} [by SSS-similarity criteria].
And also, {tex}\angle B = \angle Q{/tex}          [corresponding angles of similar triangles are equal].
Now, in {tex}\triangle ABC{/tex} and {tex}\triangle PQR{/tex}, we have
{tex}\angle B = \angle Q{/tex}   [proved above]
and {tex}\frac { A B } { P Q } = \frac { B D } { Q M }{/tex} [from(i)].
{tex}\therefore \quad \triangle A B C \sim \triangle P Q R{/tex}    [by SAS-similarity criteria].