So ∠ABC=∠BCD

Because < 1 = < 2.

So ∠EBC=∠BCF

So parallel to CF

2.< 1 = < 2 = 60 ed bisector BEF

Bed = 60.

∠BEG= 180 -∠BED-∠2=60

∠BEG=∠ 1

So AB//CD

3.∫de divides ∠CDA ∴ ∠ADE=∠EDC = 1/2∠ADC.

Similarly ∠FBA∠CBF= 1/2∠ABC.

∠∠CDA =∠CBA∴∠ade =∠EDC =∠CBF =∠FBA

∠∠ade =∠aed∴∠FBA =∠ade =∠aed

∴ de BF (same angle, two parallel straight lines)