If AD and PM are medians of ∆s ABC and PQR respectively where ∆ABC~∆PQR prove that AB/PQ =AD/PM

If AD and PM are medians of ∆s ABC and PQR respectively where ∆ABC

1.	If AD and PM are medians of ∆s ABC and PQR respectively where ∆ABC~∆PQR prove that AB/PQ =AD/PM
Answer» Given: AD and PM are median of triangles ABC PQR respectively where {tex}\\triangle {/tex} ABC {tex} \\sim {/tex}{tex}\\triangle {/tex}PQRTo prove: {tex}\\frac{{AB}}{{PQ}} = \\frac{{AD}}{{PM}}{/tex}Proof: {tex}\\triangle {/tex}ABC {tex} \\sim {/tex}{tex}\\triangle {/tex}PQR ........Given{tex}\\therefore {/tex}{tex}\\frac{{AB}}{{PQ}} = \\frac{{BC}}{{QR}} = \\frac{{CA}}{{RP}}{/tex} .......(1).....[{tex}\\because {/tex} Corresponding sides of two similar triangles are proportional]and{tex}\\angle{/tex} A = {tex}\\angle{/tex} P, {tex}\\angle{/tex} B = {tex}\\angle{/tex} Q, {tex}\\angle{/tex} C ={tex}\\angle{/tex} R, ..........(2) [ {tex}\\because {/tex} corresponding sides of two similar triangles are proportional]But BC = 2BD and QR = 2QM.............. {tex}\\because {/tex} AD and PM are mediansSo, from(1), {tex}\\frac{{AB}}{{PQ}} = \\frac{{2BD}}{{2QM}}{/tex}{tex}\\Rightarrow {/tex}{tex}\\frac{{AB}}{{PQ}} = \\frac{{BD}}{{QM}}{/tex} ........(3)Also, {tex}\\angle{/tex} ABD = {tex}\\angle{/tex} PQM .........(4).......... From (2){tex}\\therefore {/tex}{tex}\\triangle {/tex} ABD {tex} \\sim {/tex}{tex}\\triangle {/tex}PQM .......SAS similarity criterion{tex}\\therefore {/tex}{tex}\\frac{{AB}}{{PQ}} = \\frac{{AD}}{{PM}}{/tex} ........{tex}\\because {/tex}[Corresponding sides of two similar triangles are proportional]