Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider a quadrilateral with vertices \(A, B, C,\) and \(D\) in order (as shown in the diagram). If the diagonals \(A C\) and \(B D\) bisect each other, show that the quadrilateral is a parallelogram. (This is the converse of Example \(4.1 .2 .)\) [Hint: Let \(E\) be the intersection of the diagonals. Show that \(\overrightarrow{A B}=\overrightarrow{D C}\) by writing \(\overrightarrow{A B}=\overrightarrow{A E}+\overrightarrow{E B} \cdot]\)

Short Answer

Expert verified
The quadrilateral is a parallelogram because \(\overrightarrow{AB} = \overrightarrow{DC}\).

Step by step solution

01

Understanding the Given

We have a quadrilateral with vertices \(A, B, C, D\). The diagonals \(AC\) and \(BD\) intersect at point \(E\) and are given to bisect each other.
02

Definition of a Parallelogram

A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel. We need to prove that \(\overrightarrow{AB} = \overrightarrow{DC}\).
03

Expressing Vectors in Terms of Intersection Point

Use the hint to express the vectors: \(\overrightarrow{AB} = \overrightarrow{AE} + \overrightarrow{EB}\) and \(\overrightarrow{DC} = \overrightarrow{DE} + \overrightarrow{EC}\).
04

Using Bisected Diagonals

Since the diagonals bisect each other, \(E\) is the midpoint of both diagonals. Therefore, \(\overrightarrow{AE} = \overrightarrow{EC}\) and \(\overrightarrow{BE} = \overrightarrow{ED}\).
05

Substitute Midpoint Equations

Substitute \(\overrightarrow{AE} = \overrightarrow{EC}\) and \(\overrightarrow{BE} = \overrightarrow{ED}\) into the expressions for \(\overrightarrow{AB}\) and \(\overrightarrow{DC}\) to show: \(\overrightarrow{AB} = \overrightarrow{EC} + \overrightarrow{DE} = \overrightarrow{DC}\).
06

Conclusion

Since \(\overrightarrow{AB} = \overrightarrow{DC}\) both in length and direction, the quadrilateral \(ABCD\) is a parallelogram.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadrilateral
A quadrilateral is a flat shape in a plane with four straight sides. Think of it as any four-sided figure formed by connecting four points, no three of which are in a straight line. When naming a quadrilateral, we often use the vertices in alphabetical order, such as \(A, B, C, D\). These vertices are connected in order to form sides \(AB, BC, CD,\) and \(DA\).
What makes quadrilaterals special is that they encompass a range of specific figures, like parallelograms, rectangles, and trapezoids, depending on certain properties. For example:
  • A parallelogram is a quadrilateral where opposite sides are equal and parallel.
  • A rectangle is a quadrilateral with all angles equal to \(90^\circ\).
  • A trapezoid only requires one pair of opposite sides to be parallel.
Identifying whether a quadrilateral is a special type often depends on its sides, angles, or how its diagonals behave. This brings us to the next important property of some quadrilaterals: how their diagonals behave.
Diagonals bisect each other
In a quadrilateral, diagonals are the line segments that connect opposite vertices. For instance, in quadrilateral \(ABCD\), the diagonals are \(AC\) and \(BD\). Sometimes, these diagonals can bisect each other. This means that they intersect at a point that divides each diagonal into two segments of equal length.
Bisecting diagonals are a key feature that hint towards specific types of quadrilaterals. If you find that the diagonals of a quadrilateral bisect each other, you can conclude that the quadrilateral must be a parallelogram. This occurs because the point of intersection of the diagonals acts as a balance point, ensuring symmetry on both sides.
Here's an easy way to visualize it: if \(E\) is the midpoint of both diagonals, you have \(AE = EC\) and \(BE = ED\). This symmetry suggests that opposite sides of the quadrilateral are equal and parallel, one of the defining properties of a parallelogram.
Vector addition
Vector addition is a mathematical way of combining two or more vectors to get a resultant vector. Vectors are represented as arrows, implying both direction and magnitude. In the context of our problem, we're dealing with vectors like \(\overrightarrow{AE}\) and \(\overrightarrow{EB}\), which are parts of the diagonals.
The rule for vector addition is simple: if you want to add vectors \(\overrightarrow{p}\) and \(\overrightarrow{q}\), you align them such that the tail of \(\overrightarrow{q}\) starts where the head of \(\overrightarrow{p}\) ends. The resultant vector \(\overrightarrow{r}\) (\(r = \overrightarrow{p} + \overrightarrow{q}\)) points from the start of \(\overrightarrow{p}\) to the end of \(\overrightarrow{q}\).
Let's tie that back to the quadrilateral. Using vector addition, we can express \(\overrightarrow{AB}\) as \(\overrightarrow{AE} + \overrightarrow{EB}\), and similarly, \(\overrightarrow{DC}\) as \(\overrightarrow{DE} + \overrightarrow{EC}\). By substituting these midpoint properties, we determine that \(\overrightarrow{AB} = \overrightarrow{DC}\), identifying one pair of equal opposite sides, confirming a parallelogram.
Midpoint
A midpoint is a point that divides a line segment into two equal halves. In our quadrilateral problem, the diagonals \(AC\) and \(BD\) intersect at point \(E\), which is the midpoint of both diagonals. This means \(E\) divides them into segments \(AE = EC\) and \(BE = ED\).
In geometric problems, the concept of midpoints can often simplify the understanding of symmetry and congruence. When diagonals intersect at their midpoints, it indicates that opposite sides are likely equal, a hint that the quadrilateral could be a parallelogram.
To find a midpoint with coordinates, say for a diagonal with endpoints \((x_1, y_1)\) and \((x_2, y_2)\), the formula to find the midpoint \((x_m, y_m)\) is given by:
  • \[x_m = \frac{x_1 + x_2}{2}\]
  • \[y_m = \frac{y_1 + y_2}{2}\]
The midpoint acts as a balancing point, ensuring that each half of the diagonal is equal. In our exercise, knowing \(E\) is the midpoint for both diagonals confirms the bisection property, pivotal in proving the quadrilateral is a parallelogram.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(\mathbf{u}\) and \(\mathbf{v}\) be nonzero, nonorthogonal vectors. If \(\theta\) is the angle between them, show that \(\tan \theta=\frac{\|\mathbf{u} \times \mathbf{v}\|}{\mathbf{u} \cdot \mathbf{v}}\).

Find the angle between the following pairs of vectors. a. \(\mathbf{u}=\left[\begin{array}{l}1 \\ 0 \\ 3\end{array}\right], \mathbf{v}=\left[\begin{array}{l}2 \\ 0 \\ 1\end{array}\right]\) b. \(\mathbf{u}=\left[\begin{array}{r}3 \\ -1 \\ 0\end{array}\right], \mathbf{v}=\left[\begin{array}{r}-6 \\ 2 \\ 0\end{array}\right]\) c. \(\mathbf{u}=\left[\begin{array}{r}7 \\ -1 \\ 3\end{array}\right], \mathbf{v}=\left[\begin{array}{r}1 \\ 4 \\ -1\end{array}\right]\) d. \(\mathbf{u}=\left[\begin{array}{r}2 \\ 1 \\ -1\end{array}\right], \mathbf{v}=\left[\begin{array}{l}3 \\ 6 \\ 3\end{array}\right]\) e. \(\mathbf{u}=\left[\begin{array}{r}1 \\ -1 \\ 0\end{array}\right], \mathbf{v}=\left[\begin{array}{l}0 \\ 1 \\ 1\end{array}\right]\) f. \(\mathbf{u}=\left[\begin{array}{l}0 \\ 3 \\ 4\end{array}\right], \mathbf{v}=\left[\begin{array}{r}5 \sqrt{2} \\ -7 \\ -1\end{array}\right]\)

a. Does the line through \(P(1,2,-3)\) with direction vector \(\mathbf{d}=\left[\begin{array}{r}1 \\ 2 \\ -3\end{array}\right]\) lie in the plane \(2 x-y-z=3 ?\) Explain. b. Does the plane through \(P(4,0,5), Q(2,2,1),\) and \(R(1,-1,2)\) pass through the origin? Explain.

In each case show that that \(T\) is either projection on a line, reflection in a line, or rotation through an angle, and find the line or angle. a. \(T\left[\begin{array}{l}x \\\ y\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}x+2 y \\ 2 x+4 y\end{array}\right]\) b. \(T\left[\begin{array}{l}x \\\ y\end{array}\right]=\frac{1}{2}\left[\begin{array}{l}x-y \\\ y-x\end{array}\right]\) c. \(T\left[\begin{array}{l}x \\\ y\end{array}\right]=\frac{1}{\sqrt{2}}\left[\begin{array}{c}-x-y \\\ x-y\end{array}\right]\) d. \(T\left[\begin{array}{l}x \\\ y\end{array}\right]=\frac{1}{5}\left[\begin{array}{c}-3 x+4 y \\ 4 x+3 y\end{array}\right]\) e. \(T\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}-y \\\ -x\end{array}\right]\) f. \(T\left[\begin{array}{l}x \\\ y\end{array}\right]=\frac{1}{2}\left[\begin{array}{l}x-\sqrt{3} y \\ \sqrt{3} x+y\end{array}\right]\)

In each case, compute the projection of \(\mathbf{u}\) on \(\mathbf{v}\) a. \(\mathbf{u}=\left[\begin{array}{l}5 \\ 7 \\ 1\end{array}\right], \mathbf{v}=\left[\begin{array}{r}2 \\ -1 \\ 3\end{array}\right]\) b. \(\mathbf{u}=\left[\begin{array}{r}3 \\ -2 \\ 1\end{array}\right], \mathbf{v}=\left[\begin{array}{l}4 \\ 1 \\ 1\end{array}\right]\) c. \(\mathbf{u}=\left[\begin{array}{r}1 \\ -1 \\ 2\end{array}\right], \mathbf{v}=\left[\begin{array}{r}3 \\ -1 \\ 1\end{array}\right]\) d. \(\mathbf{u}=\left[\begin{array}{r}3 \\ -2 \\ -1\end{array}\right], \mathbf{v}=\left[\begin{array}{r}-6 \\ 4 \\ 2\end{array}\right]\)

See all solutions

Recommended explanations on Economics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free