_________________________Given — EF || CD∠ ABC = 40°∠ CEF = 165°∠ BCE = 24°________________________________To prove — AB || CD________________________________Construction — Draw EO perpendicular to CD.Hence, ∠ COE = 90° ( —1And, ∠ OEF = 90° ( —2________________________________Also, ∠ CEO + ∠ OEF = ∠ CEF( PUTTING values, we get )Or, ∠ CEO + 90° = 165° ( From — 2 , Also, ∠ CEO And ∠ OEF are adjacent angles )Or, ∠ CEO = 165° — 90°Or, ∠ CEO = 75°________________________________Now, in TRIANGLE, CEO,∠ CEO + ∠ COE + ∠ OCE = 180° { Angle sum PROPERTY of a triangle }Or, 75° + 90° + ∠ OCE = 180° { ∠ CEO = 75° — Proved above }Or, ∠ OCE = 180° – 75° + 90°Or, ∠ OCE = 180° – 165°Or, ∠ OCE = 15 °Or, ∠ DCE = 15 °________________________________Now, ∠ BCE + ∠ DCE = ∠ BCDOr, 25° + 15° = ∠ BCDOr, ∠ BCD = 40°________________________________Here, ∠ BCD = ∠ ABC = 40°Also, ∠ BCD and ∠ ABC are alternate INTERIOR angles.( Theorem — If the pair of alternate interior angles are equal then the lines are parallel. )Hence, AB || CD — Proved ________________________________