If two triangles are equiangular,prove that …

Given: Two triangles ABC and DEF in which {tex}\angle{/tex} A = {tex}\angle{/tex}D, {tex}\angle{/tex}B = {tex}\angle{/tex}E and {tex}\angle{/tex}C = {tex}\angle{/tex}F, AL and DM are angle bisectors of {tex}\angle{/tex}A
and {tex}\angle{/tex}D respectively
To prove: {tex}\frac{{BC}}{{EF}} = \frac{{AL}}{{DM}}{/tex}
Proof: Triangle ABC and DEF are Similar.
{tex}\Rightarrow {/tex} {tex}\frac{{AB}}{{DE}} = \frac{{BC}}{{EF}}{/tex} ......(i)
In ​{tex}\triangle {/tex}​ ABL and ​{tex}\triangle {/tex}​ DEM, we have
{tex}\angle{/tex}B= {tex}\angle{/tex}E [Given]
{tex}\angle{/tex} BAL= {tex}\angle{/tex} EDM [ {tex}\because {/tex} {tex}\angle{/tex} A= {tex}\angle{/tex} D {tex}\Rightarrow {/tex} {tex}\frac{1}{2}\angle A = \frac{1}{2}\angle D{/tex}] 
{tex}\Rightarrow {/tex} ​{tex}\triangle {/tex}​ ABL {tex} \sim {/tex} ​{tex}\triangle {/tex}​ DEM [AA similarity]
{tex}\Rightarrow {/tex} {tex}\frac{{AB}}{{DE}} = \frac{{AL}}{{DM}}{/tex} .......(ii)
From (i) and (ii) we have
{tex}\frac{{BC}}{{EF}} = \frac{{AL}}{{DM}}{/tex}