Sides AB and AC and median …

Given : In {tex}\Delta A B C \text { and } \Delta P Q R{/tex}  The AD and PM are their medians,
such that {tex}\frac { A B } { P Q } = \frac { A D } { P M } = \frac { A C } { P R }{/tex}
To prove : {tex}\Delta A B C \sim \Delta P Q R{/tex}
Construction : Produce AD to E such that AD = DE and produce PM to N such that PM = MN. Join CE and RN.
Proof : In {tex}\Delta A B D \text { and } \Delta E D C{/tex} 
{tex}AD=DE{/tex} 
{tex}\angle A D B = \angle E D C{/tex} (vertically opposite angles)
{tex}BD=DC\text{(as AD is a median)}{/tex}  
{tex}\therefore \quad \Delta A B D \equiv \Delta E D C{/tex} (By SAS congruency)
or, {tex}AB=CE{/tex} (By CPCT)
Similarly, PQ = RN  {tex}{/tex}
{tex}\frac { A B } { P Q } = \frac { A D } { P M } = \frac { A C } { P R }{/tex} (Given)
or, {tex}\frac { C E } { R N } = \frac { 2 A D } { 2 P M } = \frac { A C } { P R }{/tex}
or {tex}\frac{CE}{RN}=\frac{AE}{PN}=\frac{AC}{PR}{/tex}
So {tex}∆ACE \sim ∆PRN{/tex} 
{tex}\angle 3=\angle 4{/tex}
Similarly {tex}\angle 1=\angle 2{/tex} 
{tex}\angle 1+\angle3=\angle2+\angle4{/tex}
So {tex}\angle A=\angle P\text{ and}{/tex} 
{tex}\frac{AB}{PQ}=\frac{AC}{PR}\text{(given)} {/tex}
Hence {tex}∆ABC\sim ∆PQR{/tex}