As shown in the figure, each edge length (including the bottom edge length) of the regular triangular prism abc-a1b1c1 is 2, e and F are the midpoint of AB and a1c1 respectively, then the cosine value of the angle formed by EF and side edge C1C is () A. 55B. 255C. 12D. 2

As shown in the figure, each edge length (including the bottom edge length) of the regular triangular prism abc-a1b1c1 is 2, e and F are the midpoint of AB and a1c1 respectively, then the cosine value of the angle formed by EF and side edge C1C is () A. 55B. 255C. 12D. 2

Take the middle point G of AC and connect FG and eg. according to the meaning of the title, we can know that FG ‖ C1C, FG = C1C; eg ‖ BC, eg = BC; ≠ EFG is the angle between EF and side edge C1C, where RT △ EFG, cos ∠ EFG = 255, so select: B

If e, F and G are the midpoint of AB, Aa1 and a1c1 respectively, then the sine value of the angle between b1d and plane GEF is?

As shown in the figure, take the midpoint g of AC and connect FG and eg,
Then FG ‖ C1C, FG = C1C; eg ‖ BC, eg = 1 / 2BC,
Therefore, EFG is the angle (or complementary angle) formed by EF and C1C,
In RT △ EFG
Cos ∠ EFG = FG / Fe = 2 / radical 5 = 2 times radical 5 / 5

As shown in the figure: in the straight triangular prism abc-a1b1c1, it is known that ab = A1A, AC = BC, points D and E are the midpoint of C1C and ab respectively, and O is the intersection of A1B and Ab1. (I) verification: EC ∥ plane a1bd; (II) verification: Ab1 ⊥ plane a1bd

(1) ∵ o is the intersection of A1B and Ab1. In straight triangular abc-a1b1c1, ab = A1A, AC = BC, ∵ o is the midpoint of A1B. In △ a1ba, ∵ e is the midpoint of AB, ∵ EO is parallel to A1A, EO = A1A2, and EO is vertical to AB, ∵ D is the midpoint of C1C, ∵ DC = C1C2 = A1A2 = EO, ∵ EO ∥ DC, and EO = DC, EO is vertical to AB, ∥ quadrilateral EODC is rectangular, ∥ EC ∥ OD, and EC = OD, ⊂ OD ⊂ plane a1bd, EC ⊄ plane A1B D. (2) the EODC of the quadrilateral is rectangular, OD ⊥ OE, ≁ AC = BC, e is the midpoint of AB, EC ⊥ AB, ≁ OD ⊥ AB, ≁ OD ⊥ plane abb1a1, ≁ od vertical Ab1, ≁ AB = A1A, ≁ side abb1a1 is square, Ab1 ⊥ A1B, ≁ A1B and od are on plane a1bd, A1B ∩ od = O, ∩ Ab1 ⊥ plane a1bd

As shown in the figure, △ ABC and △ alblc1 are equilateral triangles, and the midpoint of BC and b1c1 is d. verification: Aa1 ⊥ CC1

It is proved that: connecting ad, extending Aa1, crossing DC to o, crossing C1C to e, ∵ - ada1 = 90 ° - a1dc = ∠ cdc1, ADDC = da1dc1 = 3, ∵ aa1d ∽ cc1d, ∵ a1ad = ∠ c1cd, and ∵ - AOD = ∠ Coe, ∵ ADO = ∠ CEO = 90 °, namely Aa1 ⊥ CC1