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Dear Student,

Please find below the solution to the asked query:

In $∆$ QLM and $∆$ RNM 

LM  =  NM                                                   ( Given )

QM  =  RM                                                  ( Given )

And

$\angle \mathrm{QLM} = \angle \mathrm{RNM} = 90° ( \mathrm{Given} \mathrm{ML}\perp \mathrm{PQ} \mathrm{and} \mathrm{MN} \perp \mathrm{PR} )$
So,

$∆$ QLM $\cong$$∆$ RNM                                    ( RHL rule )

So,

$\angle$ MQL  =  $\angle$ MRN                                  ( BY CPCT )

Now from above equation and from base angle theorem we can get in $∆$ PQR :

PQ  = PR                                                     ( Hence proved )


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