Question

Use vectors to prove the following theorems from geometry; The diagonals of a parallelogram bisect each other.

Short answer

Midpoints of diagonals \(\textbf{AC}\) and \(\textbf{BD}\) are equal, proving the diagonals bisect each other.

Step-by-step solution

Define the Parallelogram

Consider a parallelogram with vertices labeled as points: A, B, C, and D. Let \(\textbf{A} = \textbf{a}\), \(\textbf{B} = \textbf{b}\), \(\textbf{C} = \textbf{c}\), and \(\textbf{D} = \textbf{d}\). Assume that vectors \(\textbf{AB}\) and \(\textbf{AD}\) are the sides of the parallelogram.

Express Vectors in Terms of \(\textbf{a}\) and \(\textbf{b}\)

Since \(\textbf{AB}\) and \(\textbf{AD}\) are the sides, \(\textbf{AB} = \textbf{b} - \textbf{a}\) and \(\textbf{AD} = \textbf{d} - \textbf{a}\). Also, \(\textbf{BC}\) and \(\textbf{CD}\) are parallel to \(\textbf{AD}\) and \(\textbf{AB}\) respectively.

Express Diagonals

The diagonals of the parallelogram are \(\textbf{AC}\) and \(\textbf{BD}\). Thus, \(\textbf{AC} = \textbf{c} - \textbf{a}\) and \(\textbf{BD} = \textbf{d} - \textbf{b}\). Since \(\textbf{C} = \textbf{a} + \textbf{d} - \textbf{b}\), it implies that \(\textbf{AC} = \textbf{a} + (\textbf{d} - \textbf{b}) - \textbf{a} = \textbf{d} - \textbf{b}\). Hence, \(\textbf{AC} = \textbf{BD}\).

Calculate Midpoints of Diagonals

We need to show that the diagonals bisect each other, which means that the midpoint of \(\textbf{AC}\) is the same as the midpoint of \(\textbf{BD}\). The midpoint of \(\textbf{AC}\) is given by: \[ \text{Midpoint of } \textbf{AC} = \frac{1}{2} (\textbf{a} + \textbf{c}) \] And the midpoint of \(\textbf{BD}\) is given by: \[ \text{Midpoint of } \textbf{BD} = \frac{1}{2} (\textbf{b} + \textbf{d}) \]

Compare Midpoints

Now, express \(\textbf{c}\) and \(\textbf{d}\) using the vector properties, where \(\textbf{c} = \textbf{a} + \textbf{d} - \textbf{b}\). Thus, \[ \text{Midpoint of } \textbf{AC} = \frac{1}{2} (\textbf{a} + \textbf{c}) = \frac{1}{2} (\textbf{a} + (\textbf{a} + \textbf{d} - \textbf{b})) \] \[ = \frac{1}{2} (2\textbf{a} + \textbf{d} - \textbf{b}) = \frac{1}{2} (\textbf{b} + \textbf{d}) \] Hence, the midpoint of \(\textbf{AC}\) = midpoint of \(\textbf{BD}\).

Key concepts

vectors in geometry

Vectors are powerful tools in geometry used to represent quantities having both magnitude and direction. They are especially useful in proving geometric theorems and properties, as they allow for compact and precise arguments. In our case, we consider vectors to represent the points of a parallelogram.
For a parallelogram with vertices labeled as points A, B, C, and D, we can assign vectors to these points, such as \(\textbf{A} = \textbf{a}\) and \(\textbf{B} = \textbf{b}\). These vectors help us to express other geometric entities, like sides and diagonals, in terms of simple vector operations.
Using vectors, we can derive relationships such as the sides \(\textbf{AB} = \textbf{b} - \textbf{a}\) and \(\textbf{AD} = \textbf{d} - \textbf{a}\). These vector expressions facilitate our understanding and manipulation of geometric properties, such as those involving a parallelogram's diagonals.

parallelogram properties

A parallelogram is a quadrilateral with opposite sides that are equal and parallel. This inherent symmetry leads to intriguing properties, one of which is the fact that the diagonals of a parallelogram bisect each other.
In a parallelogram, sides and diagonals have specific relationships defined by vector additions and subtractions. For instance, in our initial setup, we express the diagonals \(\textbf{AC}\) and \(\textbf{BD}\) in terms of vector components. Utilizing the vectors, we derive these as follows: \(\textbf{AC} = \textbf{c} - \textbf{a}\) and \(\textbf{BD} = \textbf{d} - \textbf{b}\).
Because \(\textbf{AC}\) and \(\textbf{BD}\) intersect each other, the midpoints of these diagonals provide a straightforward method to show that they bisect each other, reinforcing the symmetry lying at the heart of parallelogram properties.

diagonals bisecting theorem

One remarkable property of a parallelogram is that its diagonals bisect each other. This means that each diagonal cuts the other into two equal halves at their point of intersection.
Using vector notation, we can prove this by comparing the midpoints of the diagonals. For diagonals \(\textbf{AC}\) and \(\textbf{BD}\), described using vectors, we find their midpoints. We start with:
Midpoint of \(\textbf{AC}\)
\[\text{Midpoint of } \textbf{AC} = \frac{1}{2} (\textbf{a} + \textbf{c}) \]
Then, we find the midpoint of \(\textbf{BD}\)
\[\text{Midpoint of } \textbf{BD} = \frac{1}{2} (\textbf{b} + \textbf{d}) \]
Given the expressions for points C and D in terms of other points and vectors, the midpoints of these diagonals simplify to the same vector value, confirming that the diagonals bisect each other. Thus,
\[\frac{1}{2} (\textbf{a} + \textbf{c}) = \frac{1}{2} (\textbf{b} + \textbf{d})\]

midpoints of diagonals

The midpoint of a segment that joins two points in geometry is crucial in many proofs and constructions. It is particularly important in the case of a parallelogram's diagonals, which intersect at their midpoints.
By definition, the midpoint of a segment is found by averaging the coordinates of the endpoints. For example, if segment \(\text{AC}\) has endpoints at \(\textbf{a}\) and \(\textbf{c}\), its midpoint is:
\[\text{Midpoint of } \textbf{AC} = \frac{1}{2} (\textbf{a} + \textbf{c}) \]
Similarly, the midpoint of diagonal \(\text{BD}\) can be determined. When a parallelogram's properties are invoked, particularly its side and angle relationships, these midpoints confirm the diagonals' bisecting theorem.
By carefully analyzing and expressing vector components, we show that these two midpoints align perfectly, demonstrating not only fundamental properties of vectors but also the inherent symmetry of parallelograms.