∫AB parallel to CD

∴∠ ebb1+∠ edd1= ∠ e =140 degrees.

Bf and df are bisectors of ∠ABE and ∠CDE, respectively.

Therefore, ∠FBE = 1/2∠ Abe, ∠FDE= 1/2∠CDE.

∠∠FBE+∠FDE = 1/2(∠ Abe +∠CDE)

= 1/2( 180 -∞ebb 1+ 180 -∞edd 1)

= 1/2( 180 degrees+180 degrees-140 degrees)

= 1/2*220 degrees

= 1 10 degrees

The sum of the internal angles of the quadrilateral BFDE =360 degrees.

∴∠BFD=360 degrees -∠E-∠FBE-∠FDE

=360 degrees-140 degrees-1 10 degrees

= 1 10 degrees.