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)