As shown in the figure, fold a triangular piece of paper ABC along De, and point a falls inside the quadrilateral bced (1). If ∠ a = α, find ∠ 1 + 2
∠A'EA+∠A'DA=360-(α+α)=360-2α
∠1+∠2=(180-∠A'EA)+(180-∠A'DA)=2α=2∠A
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