Here are some problems from Multivariable and Vector Calculus by David Santos.

Varignon’s Theorem: The quadrilateral formed by midpoints of sides of any quadrilateral is a parallelogram.

Proof: We proceed by vectors. Let our arbitrary quadrilateral be \(ABCD\), and let \(W\) be the midpoint of \(\overline{AB}\), \(X\) be the midpoint of \(\overline{BC}\), etc.

Observe that it suffices to show that \(\mathbf{ZW}+\mathbf{ZY}=\mathbf{ZX}\).

We see that \(\mathbf{ZW}=\mathbf{ZA}+\mathbf{AW}\). Furthermore, \(\mathbf{ZY}=\mathbf{ZD}+\mathbf{DY}=-\mathbf{ZA}+\mathbf{DY}\). Adding these two equations yields: \[\mathbf{ZW}+\mathbf{ZY}=\mathbf{AW}+\mathbf{DY}\] We take a look at \(\mathbf{AW}\) and find \(\mathbf{AW}=\mathbf{WB}\) and \(\mathbf{WB}=\mathbf{WX}-\mathbf{BX}\), hence \(\mathbf{AW}=\mathbf{WX}-\mathbf{BX}\). In a similar manner we may show that \(\mathbf{DY}=\mathbf{YX}-\mathbf{CX}=\mathbf{YX}+\mathbf{BX}\). Therefore: \[\mathbf{AW}+\mathbf{DY}=\mathbf{YX}+\mathbf{WX}\] And so: \[\mathbf{ZW}+\mathbf{ZY}=\mathbf{YX}+\mathbf{WX}\] We’re almost done. We add \(\mathbf{YX}+\mathbf{WX}\) to both sides to obtain: \[\mathbf{ZW}+\mathbf{WX}+\mathbf{ZY}+\mathbf{YX}=2\left(\mathbf{YX}+\mathbf{WX}\right)\] This simplifies to: \[2\mathbf{ZX}=2\left(\mathbf{YX}+\mathbf{WX}\right)=2\left(\mathbf{ZW}+\mathbf{ZY}\right)\] And dividing by \(2\), we obtain the desired result. \(\square\)

Problem: Let \(X\), \(Y\), and \(Z\) be points on the plane with \(X\neq Y\). Demonstrate that the point \(A\) belongs to \(\overleftrightarrow{XY}\) if and only if there exists scalars \(\alpha,\beta\) with \(\alpha+\beta=1\) such that: \[\mathbf{ZA}=\alpha\mathbf{ZX}+\beta\mathbf{ZY}\]

Solution: We define \(\overleftrightarrow{XY}\) to be the standard horizontal axis. Then, the component of any vector from \(Z\) to \(\overleftrightarrow{XY}\) orthogonal to \(\overleftrightarrow{XY}\) must be a constant. Hence we may write: \[\mathbf{ZX}=\left<a,k\right>\] \[\mathbf{ZY}=\left<b,k\right>\] Suppose \(\exists\alpha,\beta\) such that \(\mathbf{ZA}=\alpha\mathbf{ZX}+\beta\mathbf{ZY}\) and \(\alpha+\beta=1\). Then it follows that we may write: \[\mathbf{ZA}=\left<\alpha a+\beta b,(\alpha+\beta)k\right>\] But since \(\alpha+\beta=1\), this becomes: \[\mathbf{ZA}=\left<\alpha a+\beta b,k\right>\] Since the component orthogonal to \(\overleftrightarrow{XY}\) of \(\mathbf{ZA}\) is equal to the constant \(k\), \(A\) must lie on \(\overleftrightarrow{XY}\).

The converse of the above is also true. Suppose that \(A\) does lie on \(\overleftrightarrow{XY}\). Then: \[\mathbf{ZA}=\left<c,k\right>\] It is possible to uniquely determine \(\alpha,\beta\) that satisfy: \[\left\{\begin{array}{ll}\alpha a+\beta b=c\\ \alpha+\beta=1\ \end{array}\right.\] By solving the system of equations above. A unique solution exists because we are given \(X\neq Y\) and thus \(a\neq b\), so the system is independent.

Those values of \(\alpha\) and \(\beta\) permit the relation: \[\mathbf{ZA}=\alpha\mathbf{ZX}+\beta\mathbf{ZY}\] Hence, \(A\) lies on \(\overleftrightarrow{XY}\) iff \(\alpha\) and \(\beta\) exist as described. \(\square\)