It is given that ABCD is a parallelogram, so AB || DC by the definition of parallelogram. So, ∠1 ≅ ∠2 by the alternate interior angles theorem. It is also given that DC bisects ∠BDE, so ∠2 ≅ ∠3 by the definition of angle bisector. Therefore, ∠1 ≅ ∠3 by the transitive property of congruence (in which ∠1 ≅ ∠2, ∠2 ≅ ∠3, ∴ ∠1 ≅ ∠3 by transitive property of congruence)
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~Rise Above the Ordinary, Senpai
