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If you connect the midpoints of two sides of any triangle, the resulting segment is always parallel to the base and exactly half its length.
Classical Euclidean geometry requires complex angle chasing and similar triangle ratios to prove this, but vector addition proves it in three clean algebraic lines.
How can simple directed arrows turn a geometric proof into effortless algebra?

Setting Up Triangle Vectors

A vector is a geometric object with both a magnitude (length) and a direction, commonly written as an arrow pointing from an initial point to a terminal point.
Let triangle ABCABC have vertex AA as the reference origin, defining side vectors AB⃗=u\vec{AB} = \mathbf{u} and AC⃗=v\vec{AC} = \mathbf{v}.
By vector subtraction, navigating from BB to CC is equivalent to moving backward along AB⃗\vec{AB} and forward along AC⃗\vec{AC}, giving the base vector BC⃗=v−u\vec{BC} = \mathbf{v} - \mathbf{u}.
What happens when we locate the midpoints on each side using these exact same base vectors?

The Algebraic Proof

By the vector representation of midpoints, if point DD (or MM) is the midpoint of segment ABAB, its directed displacement vector is AD⃗=12AB⃗\vec{AD} = \frac{1}{2}\vec{AB}. Relative to any reference origin OO, its position vector is OD⃗=d=a+b2\vec{OD} = \mathbf{d} = \frac{\mathbf{a} + \mathbf{b}}{2}.
To find midsegment vector DE⃗\vec{DE}, we subtract the endpoint vectors: DE⃗=AE⃗−AD⃗=12AC⃗−12AB⃗\vec{DE} = \vec{AE} - \vec{AD} = \frac{1}{2}\vec{AC} - \frac{1}{2}\vec{AB}.