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As shown in the figure, ABCD is a square with side length of 2a, Pb ⊥ plane ABCD, Ma ∥ Pb, and Pb = 2mA = 2A, e is the midpoint of PD Verification: me ‖ plane ABCD Solution: cosine value of dihedral angle formed by plane PMD and plane ABCD Can you write me the specific process of the first question? I really don't understand. Please
(1) Take the midpoint o of BD and connect EO and AO
So Ma and EO are parallel and equal
So maoe is parallelogram, so me / / AO, so me / / surface ABCD
(2) The projection of △ PMD on surface ABCD is △ abd
S(△ABD)=2a^2
PM=√5a,MD=√5a,PD=2√3a
So s (△ PMD) = √ 6A ^ 2
Let the dihedral angle between PMD and ABCD be θ
Then cos θ = s (△ ABD) / S (△ PMD) = 2 / √ 6 = √ 6 / 3
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