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


Also, EF||BD,

∠1=∠2 and ∠3=∠4

So, From all the above 3 equations,




So, BE=DF.

(ii) In triangles ADF and ABE,


∠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)


AM bisects ∠BAD

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