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Draw any messy, lopsided four-sided polygon in 3D space, connect the midpoints of its four edges, and you will always get a perfect flat parallelogram.
Proving this using classic Euclidean angles is a headache, but vector algebra solves it in just three lines.

Position Vectors and the Ratio Theorem

A position vector is an arrow drawn from a fixed origin OO to any point in space, written as a⃗=OA⃗\vec{a} = \vec{OA}.
The vector displacement from point AA to point BB is always the destination position minus the starting position: ABβƒ—=bβƒ—βˆ’aβƒ—\vec{AB} = \vec{b} - \vec{a}.
If a point PP divides the line segment ABAB internally in the ratio m:nm : n, we find its position vector p⃗\vec{p} using the section formula: p⃗=na⃗+mb⃗m+n\vec{p} = \frac{n\vec{a} + m\vec{b}}{m+n}.
Why does this work? Since APβƒ—=mm+nABβƒ—\vec{AP} = \frac{m}{m+n}\vec{AB}, we have pβƒ—βˆ’aβƒ—=mm+n(bβƒ—βˆ’aβƒ—)\vec{p} - \vec{a} = \frac{m}{m+n}(\vec{b} - \vec{a}), which rearranges directly to the section formula.
For a midpoint MM, the ratio is 1:11:1, giving the familiar average m⃗=a⃗+b⃗2\vec{m} = \frac{\vec{a} + \vec{b}}{2}.

Collinearity and Linear Independence

Under the Collinearity Criterion, three distinct points AA, BB, and CC are collinear if and only if ABβƒ—=kBCβƒ—\vec{AB} = k\vec{BC} (or ACβƒ—=Ξ»ABβƒ—\vec{AC} = \lambda\vec{AB}) for some non-zero scalar k,λ∈Rk, \lambda \in \mathbb{R}.
Because the displacement vectors share point BB and point along the same line of action, a scalar multiplier proves straight-line alignment without calculating slopes.
In 2D, two non-parallel vectors exhibit vector linear independence: the linear combination αu⃗+βv⃗=0\alpha\vec{u} + \beta\vec{v} = \mathbf{0} holds if and only if α=0\alpha = 0 and β=0\beta = 0.