Vector Proof II — Classical Geometry
Once you treat points as position vectors and midpoints as averages, classical Euclidean theorems collapse into one-line algebra. In this lesson you will prove that the diagonals of a parallelogram bisect each other, that the three medians of a triangle meet at the centroid in a $1:2$ ratio, that the segment joining two midpoints is parallel to and half the length of the third side, and that the perpendicular bisectors of a triangle's sides are concurrent. These are the canonical NESA MEX-V1 vector proofs.
Points $A, B, C$ have position vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ from a fixed origin $O$. Without a coordinate system — write the position vector of (a) the midpoint $M$ of $AB$, (b) the point $P$ on $AC$ such that $AP:PC = 1:2$. Justify each with one line.
Every classical theorem in this lesson succumbs to the same two-step routine: label every point as a position vector from any chosen origin, then write the candidate intersection or midpoint as an explicit vector expression. If the expression is symmetric in the three vertices, concurrency follows for free.
The position-section-symmetry routine: (1) pick any origin (or a vertex, to kill one vector), (2) write each named point as a linear combination of $\mathbf{a}, \mathbf{b}, \mathbf{c}$ using the section formula, (3) check the resulting expression for symmetry — if it treats the three vertices the same way, the three lines must meet there.
Section formula: point dividing $PQ$ in ratio $m:n$ has position vector $\dfrac{n\mathbf{p} + m\mathbf{q}}{m+n}$.
Key facts
- Midpoint of $AB$ has position vector $\tfrac{1}{2}(\mathbf{a}+\mathbf{b})$
- Centroid of $\triangle ABC$ is $\tfrac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})$
- Centroid divides each median in ratio $2:1$ from the vertex
- $\vec{PQ} \parallel \vec{RS} \iff \vec{PQ} = k\,\vec{RS}$ for some scalar $k$
Concepts
- Why a vector identity yields both direction and length at once
- Why a symmetric expression in $\mathbf{a},\mathbf{b},\mathbf{c}$ forces concurrency
- Why the midpoint theorem follows in one line from subtraction
- Why perpendicular bisectors meet at the circumcentre
Skills
- Prove parallelogram diagonals bisect each other using vectors
- Prove the three medians of a triangle are concurrent at the centroid
- Prove the midpoint theorem $\vec{MN} = \tfrac{1}{2}\vec{BC}$
- Prove perpendicular bisectors are concurrent using the dot product
Two classical theorems each yield to a single vector calculation.
(i) Parallelogram diagonals bisect each other. Let $ABCD$ be a parallelogram with position vectors $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$. The parallelogram condition is $\vec{AB}=\vec{DC}$, equivalently $\mathbf{b}-\mathbf{a}=\mathbf{c}-\mathbf{d}$, i.e. $\mathbf{a}+\mathbf{c}=\mathbf{b}+\mathbf{d}$. Now the midpoint of diagonal $AC$ has position vector $\tfrac{1}{2}(\mathbf{a}+\mathbf{c})$ and the midpoint of $BD$ has position vector $\tfrac{1}{2}(\mathbf{b}+\mathbf{d})$. These are equal by the parallelogram condition, so the diagonals share a midpoint — they bisect each other. Conversely, if the diagonals bisect each other the same equation forces $ABCD$ to be a parallelogram.
(ii) Midpoint theorem. Let $\triangle ABC$ have $M$ the midpoint of $AB$ and $N$ the midpoint of $AC$. Then $\mathbf{m}=\tfrac{1}{2}(\mathbf{a}+\mathbf{b})$ and $\mathbf{n}=\tfrac{1}{2}(\mathbf{a}+\mathbf{c})$, so $$\vec{MN}=\mathbf{n}-\mathbf{m}=\tfrac{1}{2}(\mathbf{c}-\mathbf{b})=\tfrac{1}{2}\vec{BC}.$$ This is a single vector identity — it simultaneously proves $MN \parallel BC$ (same direction up to scalar) and $|MN|=\tfrac{1}{2}|BC|$.
Parallelogram $ABCD$: $\mathbf{a}+\mathbf{c}=\mathbf{b}+\mathbf{d}$ · Diagonals share midpoint $\tfrac{1}{2}(\mathbf{a}+\mathbf{c})=\tfrac{1}{2}(\mathbf{b}+\mathbf{d})$ ⇒ bisect · Midpoint theorem: $M=\tfrac{1}{2}(\mathbf{a}+\mathbf{b})$, $N=\tfrac{1}{2}(\mathbf{a}+\mathbf{c})$ ⇒ $\vec{MN}=\tfrac{1}{2}\vec{BC}$ · Single identity = parallel + half length, both at once
Pause — copy $\mathbf{a}+\mathbf{c} = \mathbf{b}+\mathbf{d}$ (parallelogram diagonal bisection), the midpoint-theorem derivation $\overrightarrow{MN} = \tfrac{1}{2}\overrightarrow{BC}$, and the two-in-one conclusion (parallel + half length) into your book.
Quick check: In $\triangle ABC$, $M$ is the midpoint of $AB$ and $N$ the midpoint of $AC$. Which vector identity is the cleanest single-line proof of the midpoint theorem?
We just saw that in parallelogram $ABCD$ the diagonals bisect each other because $\mathbf{a}+\mathbf{c} = \mathbf{b}+\mathbf{d}$, and the midpoint theorem gives $\overrightarrow{MN} = \tfrac{1}{2}\overrightarrow{BC}$ from one vector identity. That raises a question: can vectors prove that the three medians of a triangle are concurrent? This card answers it → the centroid $\mathbf{g} = \tfrac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})$ is symmetric in the three vertices, so it lies on all three medians in ratio $2:1$.
(iii) Three medians of $\triangle ABC$ are concurrent. Let $M_A,M_B,M_C$ be midpoints of the sides opposite $A,B,C$, so $\mathbf{m}_A=\tfrac{1}{2}(\mathbf{b}+\mathbf{c})$ etc. Consider the point $\mathbf{g}=\tfrac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})$. Rewrite this as $$\mathbf{g}=\tfrac{1}{3}\mathbf{a}+\tfrac{2}{3}\cdot\tfrac{1}{2}(\mathbf{b}+\mathbf{c})=\tfrac{1}{3}\mathbf{a}+\tfrac{2}{3}\mathbf{m}_A.$$ This shows $G$ lies on median $AM_A$, $\tfrac{2}{3}$ of the way from $A$ to $M_A$. By the symmetry of $\mathbf{g}$ in $\mathbf{a},\mathbf{b},\mathbf{c}$, the same calculation shows $G$ lies on $BM_B$ and $CM_C$ in the same ratio. Hence all three medians pass through $G$ and divide each in the ratio $\mathbf{AG:GM_A=2:1}$.
(iv) Perpendicular bisectors are concurrent. Let $P$ be the intersection of the perpendicular bisectors of $AB$ and $AC$. From the definition of perpendicular bisector, $|PA|=|PB|$ and $|PA|=|PC|$. Hence $|PB|=|PC|$, which means $P$ lies on the perpendicular bisector of $BC$. So all three perpendicular bisectors meet at $P$ — the circumcentre. Dot product form: "$P$ on perpendicular bisector of $XY$" is equivalent to $|\mathbf{p}-\mathbf{x}|^2=|\mathbf{p}-\mathbf{y}|^2$, which expands to $2(\mathbf{y}-\mathbf{x})\cdot\mathbf{p}=|\mathbf{y}|^2-|\mathbf{x}|^2$, a single linear equation in $\mathbf{p}$.
Centroid: $\mathbf{g}=\tfrac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})$ — symmetric in the three vertices · Factorisation $\mathbf{g}=\tfrac{1}{3}\mathbf{a}+\tfrac{2}{3}\mathbf{m}_A$ ⇒ on median $AM_A$ in ratio $2:1$ · Symmetry ⇒ same point lies on all three medians · Perpendicular bisectors meet at circumcentre via $|PA|=|PB|=|PC|$
Pause — copy the centroid formula $\mathbf{g} = \tfrac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})$, the $2:1$ ratio on each median, the symmetry argument for concurrence, and the circumcentre condition $|PA|=|PB|=|PC|$ into your book.
Did you get this? True or false: the centroid $G$ of $\triangle ABC$ divides the median from $A$ to the midpoint of $BC$ in the ratio $AG:GM_A = 2:1$.
Worked examples · 3 in a row, reveal as you go
$ABCD$ is a parallelogram. Using position vectors from an origin $O$, prove that the diagonals $AC$ and $BD$ bisect each other.
$\triangle ABC$ has position vectors $\mathbf{a},\mathbf{b},\mathbf{c}$ from origin $O$. Prove that the three medians meet at a single point $G$ and that $AG:GM_A=2:1$, where $M_A$ is the midpoint of $BC$.
Use position vectors and the dot product to prove that the three perpendicular bisectors of the sides of $\triangle ABC$ are concurrent at a point equidistant from $A$, $B$ and $C$.
Fill the gap: The position vector of the centroid of $\triangle ABC$ is $\mathbf{g}=\dfrac{1}{}\big(\mathbf{a}+\mathbf{b}+\mathbf{c}\big)$, and the centroid divides each median in the ratio $$ measured from the vertex.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: if $M$ and $N$ are midpoints of sides $AB$ and $AC$ of $\triangle ABC$, then $\vec{MN}=\tfrac{1}{2}\vec{BC}$ proves both that $MN \parallel BC$ and $|MN|=\tfrac{1}{2}|BC|$.
Activities · practice with the ideas
$PQRS$ is a parallelogram. Let $X$ be the midpoint of $PR$. Using position vectors, show that $X$ is also the midpoint of $QS$.
$\triangle ABC$ has $A=(1,0,0), B=(0,1,0), C=(0,0,1)$. Find the centroid $G$ and verify that $AG:GM_A=2:1$, where $M_A$ is the midpoint of $BC$.
In $\triangle ABC$ let $M,N$ be midpoints of $AB,AC$. Prove $\vec{MN}=\tfrac{1}{2}\vec{BC}$ using position vectors, and explain why this proves both parallelism and the half-length relation.
Using the dot product, write the condition "$P$ is equidistant from $A$ and $B$" as a linear equation in $\mathbf{p}$. Hence describe the perpendicular bisector of $AB$ as a plane (in 3D) or a line (in 2D).
$ABCD$ is a quadrilateral. Let $P,Q,R,S$ be the midpoints of $AB,BC,CD,DA$. Prove that $PQRS$ is a parallelogram (Varignon's theorem).
Odd one out: Three of these expressions are valid position vectors for the centroid of $\triangle ABC$. Which one is NOT?
Earlier you wrote the midpoint of $BC$ as $\mathbf{m}_A=\tfrac{1}{2}(\mathbf{b}+\mathbf{c})$ and then the point two-thirds of the way along $AM_A$ from $A$.
You should have obtained $\mathbf{a}+\tfrac{2}{3}(\mathbf{m}_A-\mathbf{a})=\tfrac{1}{3}\mathbf{a}+\tfrac{2}{3}\mathbf{m}_A=\tfrac{1}{3}\mathbf{a}+\tfrac{1}{3}(\mathbf{b}+\mathbf{c})=\tfrac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})$. The striking observation is that this is symmetric in $\mathbf{a},\mathbf{b},\mathbf{c}$ — the very feature that proves the same point lies on all three medians. The single symbolic combination $\tfrac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})$ is both the centroid formula and the concurrency proof in one.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. $ABCD$ is a parallelogram with position vectors $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$. Using the parallelogram condition $\vec{AB}=\vec{DC}$, prove that the midpoints of the diagonals $AC$ and $BD$ coincide. (2 marks)
Q2. $\triangle ABC$ has position vectors $\mathbf{a},\mathbf{b},\mathbf{c}$. Let $M$ be the midpoint of $BC$ and let $G$ be the point on $AM$ with $\vec{AG}=\tfrac{2}{3}\vec{AM}$. Show that $\mathbf{g}=\tfrac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})$, and hence explain why all three medians of $\triangle ABC$ are concurrent. (3 marks)
Q3. $\triangle ABC$ has $M, N$ the midpoints of $AB, AC$. (a) Prove $\vec{MN}=\tfrac{1}{2}\vec{BC}$. (b) Hence state the two geometric conclusions that follow from this single identity. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. Parallelogram: $\mathbf{p}+\mathbf{r}=\mathbf{q}+\mathbf{s}$ (since $\vec{PQ}=\vec{SR}$). Midpoint of $PR$ is $\tfrac{1}{2}(\mathbf{p}+\mathbf{r})$; midpoint of $QS$ is $\tfrac{1}{2}(\mathbf{q}+\mathbf{s})$. These are equal, so $X$ is the common midpoint.
2. $G=\tfrac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})=(\tfrac{1}{3},\tfrac{1}{3},\tfrac{1}{3})$. $M_A=\tfrac{1}{2}(\mathbf{b}+\mathbf{c})=(0,\tfrac{1}{2},\tfrac{1}{2})$. $\vec{AG}=(-\tfrac{2}{3},\tfrac{1}{3},\tfrac{1}{3})$, $\vec{GM_A}=(-\tfrac{1}{3},\tfrac{1}{6},\tfrac{1}{6})=\tfrac{1}{2}\vec{AG}$. Hence $AG:GM_A=2:1$.
3. $\mathbf{m}=\tfrac{1}{2}(\mathbf{a}+\mathbf{b})$, $\mathbf{n}=\tfrac{1}{2}(\mathbf{a}+\mathbf{c})$, so $\vec{MN}=\mathbf{n}-\mathbf{m}=\tfrac{1}{2}(\mathbf{c}-\mathbf{b})=\tfrac{1}{2}\vec{BC}$. The identity says $\vec{MN}$ is a scalar multiple of $\vec{BC}$ — directions parallel; and the scalar is $\tfrac{1}{2}$ — magnitudes scaled by half. Both conclusions in one line.
4. $|\mathbf{p}-\mathbf{a}|^2=|\mathbf{p}-\mathbf{b}|^2$ expands to $2\mathbf{p}\cdot(\mathbf{b}-\mathbf{a})=|\mathbf{b}|^2-|\mathbf{a}|^2$. This is a linear equation; the set of solutions is a plane in 3D (line in 2D) with normal $\mathbf{b}-\mathbf{a}$ through the midpoint $\tfrac{1}{2}(\mathbf{a}+\mathbf{b})$.
5. $\vec{PQ}=\mathbf{q}-\mathbf{p}=\tfrac{1}{2}(\mathbf{b}+\mathbf{c})-\tfrac{1}{2}(\mathbf{a}+\mathbf{b})=\tfrac{1}{2}(\mathbf{c}-\mathbf{a})$. Similarly $\vec{SR}=\mathbf{r}-\mathbf{s}=\tfrac{1}{2}(\mathbf{c}+\mathbf{d})-\tfrac{1}{2}(\mathbf{a}+\mathbf{d})=\tfrac{1}{2}(\mathbf{c}-\mathbf{a})$. Hence $\vec{PQ}=\vec{SR}$, so $PQRS$ is a parallelogram.
Q1 (2 marks): $\vec{AB}=\vec{DC}$ gives $\mathbf{b}-\mathbf{a}=\mathbf{c}-\mathbf{d}$, i.e. $\mathbf{a}+\mathbf{c}=\mathbf{b}+\mathbf{d}$ [1]. Midpoint of $AC$ is $\tfrac{1}{2}(\mathbf{a}+\mathbf{c})$, midpoint of $BD$ is $\tfrac{1}{2}(\mathbf{b}+\mathbf{d})$; these are equal, so the diagonals share a midpoint and bisect each other [1].
Q2 (3 marks): $\mathbf{m}=\tfrac{1}{2}(\mathbf{b}+\mathbf{c})$ [1]. $\mathbf{g}=\mathbf{a}+\tfrac{2}{3}(\mathbf{m}-\mathbf{a})=\tfrac{1}{3}\mathbf{a}+\tfrac{2}{3}\mathbf{m}=\tfrac{1}{3}\mathbf{a}+\tfrac{1}{3}(\mathbf{b}+\mathbf{c})=\tfrac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})$ [1]. The formula is symmetric in $\mathbf{a},\mathbf{b},\mathbf{c}$, so an identical calculation starting from $B$ or $C$ produces the same $G$. Hence the three medians all pass through $G$ — concurrent [1].
Q3 (3 marks): (a) $\mathbf{m}=\tfrac{1}{2}(\mathbf{a}+\mathbf{b})$, $\mathbf{n}=\tfrac{1}{2}(\mathbf{a}+\mathbf{c})$, so $\vec{MN}=\mathbf{n}-\mathbf{m}=\tfrac{1}{2}(\mathbf{c}-\mathbf{b})=\tfrac{1}{2}\vec{BC}$ [1]. (b) Conclusions: (i) $\vec{MN}$ is a scalar multiple of $\vec{BC}$ ⇒ $MN \parallel BC$ [1]; (ii) the scalar is $\tfrac{1}{2}$ ⇒ $|MN|=\tfrac{1}{2}|BC|$ [1].
Five timed questions on parallelogram diagonals, the centroid, the midpoint theorem and perpendicular bisectors. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
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