# in a parallelogram ABCD , m and n are the mid points of sides AB and CD respectively . prove that triangle ADN is congruent to triangle CBM

Dear Student,

We form our diagram from given information , As : Here , AB  =  CD and BC  =  DA                                                        ------ ( 1 )

And

AM  =  BM = $\frac{\mathrm{AB}}{2}$ and CN  =  DN   = $\frac{\mathrm{CD}}{2}$                                   ------ ( 2 )

From equation 1 and 2 we get

AM  =  BM  =  CN  =  DN                                                                 ------- ( 3 )

Now in $∆$ ADN  and $∆$ CBM

BC =  DA                                                                ( From equation 1 )

$\angle$ ADN  =  $\angle$ CBM                                               ( We know opposite angles in parallelogram are equal )

And

DN  =  BM                                                              ( From equation 3 )

So,

$∆$ ADN  $\cong$ $∆$ CBM                                             (  By SAS rule )                            ( Hence proved )

Hope this information will clear your doubts about Congruence of Triangles .

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Regards

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