ABCD is a square and EF is parallel to diagonal BD and EM=FM. prove that

DF=BE

AM bisects angle BAD

(i)Since diagonal of a square bisects the vertex and BD is the diagonal of square ABCD

∴     CBD =   CDB = 90 2 = 45 " src="https://s3mn.mnimgs.com/img/shared/editlive_temp/mathmlequation8289848564628255736.png" style="margin: 0px; padding: 0px; border: 0px; max-width: 550px;" /

Given :EF || BD

⇒ CEF = CBD = 45 and CEF = CDB = 45 (corresponding angles)

⇒ CEF = CFE

⇒ CE = CF (sides opposite of equal angles are equal) .......(1)

Now BC = CD (Sides of square) ........(2)

Subtracting (1) from (2) we get

⇒ BC CE = CD CF

⇒ BE = DF

(ii)∆ABE ≅ ∆ADF (by SAS congruency criterion)

⇒ BAE = DAF and AE = AF ........(3)

and ∆AER ≅ ∆AFR (by SSS congruency criterion)

⇒ EAR = FAR .........(4)

Now Adding (3) and (4) we get

⇒ BAE + EAR = DAF + FAR

⇒ BAR = DAR

i.e AR bisects BAD