It is known that the side edges and the bottom sides of the triangular prism abc-a1b1c1 are equal, and the projection D of A1 on the bottom ABC is the midpoint of BC, then the cosine value of the angle between AB and CC1 is? Trouble with space vector and solid geometry method to give solutions

Let the projection of A1 on the bottom ABC be D and the length of the side edge be X
Make de ⊥ AB through D, cross AB to e and connect a1e
Ad = xsin60 ° = radical 3 / 2 * x
A1d = under root [x ^ 2 - (root 3 / 2 * x) ^ 2] = x / 2
De = x / 2 * sin60 ° = radical 3 / 4 * x
In right triangle a1de: a1e = under root [(x / 2) ^ 2 + (root 3 / 4 * x) ^ 2] = root 7 / 4 * x
AE = under root [x ^ 2 - (root 7 / 4 * x) ^ 2] = 3 / 4 * x
cos∠A1ADB=AE/A1A=3/4*X/X=3/4

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