Proving Geometry with Vectors

Vectors turn geometry into algebra. Fix an origin O, and every point A gets a position vector \vec{a} = \vec{OA}. The single most useful fact follows at once: the vector from A to B is the difference of their position vectors,

\vec{AB} = \vec{b} - \vec{a},

because to go from A to B you can travel back to O along -\vec{a} and then out to B along \vec{b}. With just this — plus adding and scaling vectors — you can prove facts that are fiddly with pure geometry.

Three tools

Almost every vector proof is built from three little translations of geometry into vector language:

Midpoint. The midpoint M of AB has position vector \vec{m} = \tfrac{1}{2}(\vec{a} + \vec{b}) — the average of the ends.
Parallel. Two segments are parallel exactly when one vector is a scalar multiple of the other: \vec{PQ} = k\,\vec{RS} for some number k \neq 0.
Collinear. Three points A, B, C lie on one straight line when \vec{AB} = k\,\vec{AC} — parallel vectors that also share the point A.

"Parallel" and "collinear" look identical in algebra — both are "one is a multiple of the other". The only difference is that collinear vectors start from a common point, so being parallel and sharing an endpoint forces the three points onto a single line.

A worked proof: the midpoint theorem

Here is a classic result made easy: the segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length. Take triangle OAB, and let M, N be the midpoints of OA and OB. Step through the argument:

The whole proof is three lines of algebra:

\vec{m} = \tfrac{1}{2}\vec{a}, \qquad \vec{n} = \tfrac{1}{2}\vec{b},

\vec{MN} = \vec{n} - \vec{m} = \tfrac{1}{2}(\vec{b} - \vec{a}) = \tfrac{1}{2}\vec{AB}.

Because \vec{MN} is a scalar multiple (\tfrac{1}{2}) of \vec{AB}, the two segments are parallel; and that scalar being \tfrac{1}{2} says MN is half the length of AB. Two geometric conclusions, read straight off one vector equation.

Proving points collinear

To show three points lie on a line, express two vectors between them and check that one is a multiple of the other. Say \vec{AB} = 2\vec{p} and, after simplifying, \vec{AC} = 3\vec{p}. Then \vec{AC} = \tfrac{3}{2}\vec{AB} — a scalar multiple sharing the point A — so A, B, C are collinear, and C is \tfrac{3}{2} of the way along the line from A through B.

Let a parallelogram have vertices O, A, C, B with \vec{OA} = \vec{a} and \vec{OB} = \vec{b}, so the fourth vertex is \vec{c} = \vec{a} + \vec{b}. The midpoint of diagonal OC is \tfrac{1}{2}(\vec{a} + \vec{b}). The midpoint of the other diagonal AB is \tfrac{1}{2}(\vec{a} + \vec{b}) as well — the same point. So the diagonals cross at their common midpoint: they bisect each other. One line of vectors, a theorem proved.

\vec{AB} = \vec{b} - \vec{a}, tip minus tail — not \vec{a} - \vec{b}. Getting the order backwards flips the vector and wrecks the proof.
A scalar multiple only proves segments parallel. To conclude points are collinear you also need them to share a point — parallel alone allows two separate parallel lines.