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,

Please find below the solution to the asked query:

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 .

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards