Question

In fig. 15, ABCD is a square and EF is parallel to diagonal BD and EM=FM. Prove that (i) DF=BE (ii) AM bisects angle BAD

Answer 1

In ΔBCD 
BC=CD

∠2 = ∠4
 EF II BD
∠1=∠2
∠3=∠4
so, ∠3=∠4

and CE=CF
or BE=DF

2) In ΔADF and ΔABE
AD=AB
∠D=∠B
as
DF=DB
so, both triangles are congurent
AF=AE and FM=EM
so, triangles AMF and AME are also congurent
 ∠7=∠8

∠7+∠5 = ∠8 + ∠6
∠MAD = ∠MAB
AM bisect ∠BAD

Answer 2

See diagram.

   ΔCFE and ΔCDB are similar as their corresponding sides CF || CD, CE || CB and EF || BD (given).

   CE / CB = CF / CD = FM / EM 
    =>  CE = CF  ---(1)

BE = BC - CE
DF = DC - CF
      = BC - CE        by  (1)
=> DF  = BE
==============
(ii)

Draw perpendiculars from M onto AB, BC, CE and DA.

ΔCMF and ΔCME are similar and congruent, as :  CE = CF, CM is common, and FM = EM.   Then the altitudes from M,   MI = MH.

Hence,  MG = GI - MI = BC - MI
              MJ = HJ - MH = CD - MH

=> MG = MJ

As  G and J are on the sides of ∠BAD, and MG = MJ, M lies on the angular bisector of ∠BAD.