ABCD is a square and EFis parallel to BD M is mid point of EF prove that AM bisects angle BAD

(i) In triangle BCD, we have  BC=CD

∠2=∠4

Also, EF||BD,

∠1=∠2 and ∠3=∠4

So, From all the above 3 equations,

∠1=∠3

So, CE=CF

BC-BE=CD-DF

So, BE=DF.

(ii) In triangles ADF and ABE,

AD=AB 

∠ADF=∠ABE (each angle is 90 degrees)

DF = BE (from (i))

So,  triangles ADF and ABE are Congruent. (By SAS)

So AF=AE and FM=EM (given), AM=AM (common)

So triangles AMF and AME are congruent (By SSS)

So ∠7=∠8.

∠7+∠5=∠8+∠6  (because ∠5=∠6)

∠MAD=∠MAB

AM bisects ∠BAD