As shown in the figure, △ ABC is an equilateral triangle, BD ⊥ plane ABC, BD ∥ CE, and CE = CA = 2bd, M is the midpoint of EA, find plane DEA ⊥ plane ECA

Take the midpoint n of AC, connect DM, Mn, BN, then Mn / / EC, Mn = 1 / 2ec
So Mn / / is equal to BD, and because BD ⊥ plane ABC, so rectangular mnbd
So DM ⊥ Mn, DM ⊥ AC (because n is the midpoint of AC, BN ⊥ AC, DM / / BN)
So DM ⊥ plane AEC, DM is in plane DEA, so plane DEA ⊥ plane ECA

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