# 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

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 $\u2206$ ADN and $\u2206$ 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,

**$\u2206$ ADN $\cong $ $\u2206$ CBM ( By SAS rule ) ( Hence proved )**

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

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