In the straight triangular prism abc-a1b1c1, if ∠ BAC = 90 ° AB = AC = Aa1, then the angle between the out of plane straight line BA1 and AC1 is equal to () A. 30°B. 45°C. 60°D. 90°

Extend CA to D, so that ad = AC, then ada1c1 is a parallelogram, ∠ da1b is the angle formed by the out of plane straight line BA1 and AC1, and a1d = A1B = DB = 2Ab, then the triangle a1db is an equilateral triangle, ∧ da1b = 60 ° so select C

In triangle ABC, it is known that BD, CE are the middle line of triangle, and BD = CE, triangle ABC is isosceles triangle

Let BD and CE intersect at point O and connect De, then E is parallel to BC, and ED = BC / 2. In △ EOD and △ cob, the congruent △ cob 〉 OE = OC / 2 = EC / 3, OD = ob / 2 = BD / 3, OD = OE is ob = OC

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