Varignon's Theorem
The midpoints of any quadrilateral form a parallelogram with half its area, by one line of vector algebra.
1 Statement
Theorem 1.1 (Varignon). Let \(A\), \(B\), \(C\), \(D\) be any four points in \(\R ^n\), and let \(K\), \(L\), \(M\), \(N\) be the midpoints of \(AB\), \(BC\), \(CD\), \(DA\). Then:
- (a)
- \(\overrightarrow {KL} = \overrightarrow {NM} = \frac 12 \overrightarrow {AC}\) and \(\overrightarrow {KN} = \overrightarrow {LM} = \frac 12 \overrightarrow {BD}\), so \(KLMN\) is a parallelogram, possibly degenerate;
- (b)
- its diagonals bisect each other at the centroid \(G = \frac 14 (A + B + C + D)\);
- (c)
- its perimeter is \(AC + BD\);
- (d)
- if \(ABCD\) is a simple quadrilateral in the plane, the area of \(KLMN\) is half the area of \(ABCD\).
The quadrilateral may be convex, non-convex or self-intersecting, and in (a)–(c) the four points need not even lie in a plane.
2 Setting
We identify each point with its position vector, so that \(\overrightarrow {PQ} = Q - P\) and the midpoint of \(PQ\) is \(\frac 12 (P + Q)\). A parallelogram is a quadrilateral \(KLMN\) with \(L - K = M - N\); equivalently, \(KL\) and \(NM\) are equal and parallel.
For vectors \(u = (u_1, u_2)\) and \(v = (v_1, v_2)\) in \(\R ^2\), write
This is bilinear and antisymmetric (\(v \wedge u = -u \wedge v\), so \(u \wedge u = 0\)), and \(\abs {u \wedge v}\) is the area of the parallelogram spanned by \(u\) and \(v\).
Lemma 2.1 (Shoelace formula). The area of a simple polygon \(P_1 P_2 \dotsm P_m\) is \(\frac 12 \absStar{\sum _{j=1}^{m} P_j \wedge P_{j+1}}\), where \(P_{m+1} = P_1\).
We use this standard fact without proof; it follows from Green’s theorem applied to the vector field \(\frac 12(-y, x)\).
3 Idea of the proof
Every statement is linear in the four points. Each side of \(KLMN\) is the difference of two midpoints, and the terms that do not involve a diagonal cancel.
4 Proof
Proof of (a).
The same computation with \(B\) and \(D\) in place of \(A\) and \(C\) gives \(N - K = M - L = \frac 12 (D - B)\).
Proof of (b). The midpoints of the diagonals \(KM\) and \(LN\) are
Proof of (c). By (a), the four sides have lengths \(\frac 12 AC\), \(\frac 12 BD\), \(\frac 12 AC\) and \(\frac 12 BD\).
Lemma 4.1 (Area from the diagonals). For any \(A, B, C, D \in \R ^2\),
Proof. Expand the right-hand side by bilinearity and use antisymmetry:
Proof of (d). By Lemmas 2.1 and 4.1, the area of \(ABCD\) is \(\frac 12 \abs {(C - A) \wedge (D - B)}\). By (a), \(KLMN\) is the parallelogram spanned by \(L - K = \frac 12 (C - A)\) and \(N - K = \frac 12 (D - B)\), so its area is
which is half the area of \(ABCD\).
5 Remarks
Remark 5.1. The parallelogram degenerates to a segment exactly when the diagonals \(AC\) and \(BD\) are parallel, and to a point exactly when both diagonals collapse, that is, \(A = C\) and \(B = D\).