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In figure, BA || ED and BC || EF. Show that ∠ABC + ∠DEF = 180°

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∵ BA || ED

∠1 = ∠2 …(i)

(Corresponding angles)

Also BC || EF

⇒ ∠2 + ∠3 = 180° ,..(ii)

(Sum of the interior angles on the  same side of a transversal is 180°)

From (i) and (ii), we get

∠1 + ∠3 = 180°

i.e. ∠ABC + ∠DEF = 180°

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