Given a parallelepiped ABCDA1B1C1D1, O1 is the intersection point of the diagonals

Given a parallelepiped ABCDA1B1C1D1, O1 is the intersection point of the diagonals A1B1C1D1. Expand vector AO in vectors AD, AB, AA1

Solution.
Let a parallelepiped ABCDA₁B₁C₁D₁ be given, where O₁ is the intersection point of the diagonals of the upper base A₁B₁C₁D₁. To expand the vector AO₁ in vectors AD, AB, AA₁, construct O – the intersection point of the diagonals of the lower base ABCD. It is the projection of the point O₁ onto the lower base. Vector AO is equal to vector ½ ∙ АС, and vector АС is equal to the sum of vectors AB and AD according to the parallelogram rule, then vector AO is equal to vector ½ ∙ (AB + AD). In the plane of the diagonal section AA₁C₁C, the vector AO₁ is equal to the sum of the vectors AO and OO₁, but OO₁ = AA₁. We get that
vector AO₁ is equal to the sum of vectors ½ ∙ (AB + AD) and AA₁ or the sum of vectors ½ ∙ AB, ½ ∙ AD and AA₁.
Answer: vector AO₁ is equal to the sum of vectors ½ ∙ AB, ½ ∙ АD and AA₁.