In the given figure, `square`ABCD is a parallelogram. If DM `bot` AC and BN `bot` AC, then show that `squareBNDM` is a parallelogram.

Explanation

`InDeltaADMand DeltaCBN,`
`because{:{(AD=CB,("opposite sides of parallelogram")),(angleAMD=angleCNB,(each90^(@))),(angleDMA=angleBCA,("alternate interior angles")):}`
`thereforeDeltaADM~=DeltaDeltaCBA" "(A AS " therem")`
Again, in `DeltaCDNand DeltaABM," "(c.p.c.t)`
Again, in `DeltaCDN and DeltaABN,`
`because{:{(CD=AB,("opposite sides of a parallelogram")),(angleDCN=angleBAM,("alternate interior angles")),(angleCN=AM,("proved above")):}`
`thereforeDeltaCDN~=DeltaABM" "("SAS theorem")`
`impliesDN=BM" "(c.p.c.t)`
Now, DM= BN and DN = BM
`impliessquareBNDM"is a parallelogram."" "(because"pairs of opposite sides are equal")`
Alternative Method :
Construction : Join BD which intersect AC at O.
Proof : In `DeltaADM and DeltaCBN,`
`because{:{(AD=CB,("opposite sides of a parallelogram")),(angleAMD=angleCNB,(each90^(@))),(angleDAM=angleBCN,("alternate interior angles")):}`
`thereforeDeltaADM~=DeltaCBN" "(A AS)`
`thereforeAM=CN" "...(1) (c.p.c.t)`

But since `squareABCD` is a parallelogram
`{:(therefore, AO = OC , .... (2)) , ("and", DO = OB , .... (3)):}}` (diagonals of a parallelogram bisect each other)
Substacting (1) from (2), we get
AO- AM = OC - CN
`implies" "OM=ON" "...(4)`
Now, from (3) and (4), we can say that diagonals of a quadrilateral BNDM bisect each other.
Hence `squareBNDM` is a parallelogram.