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Chapter 5: Problem 159

(a) Let \(A B C D\) be a quadrilateral in which \(A B \| D C,\) and \(\underline{A B}=\underline{D C}\). Prove that \(B C \| A D,\) and hence that \(A B C D\) is a parallelogram. (b) Let \(A B C D\) be a quadrilateral in which \(\underline{A B}=\underline{D C}\) and \(\underline{B C}=\underline{A D}\). Prove that \(A B \| D C,\) and hence that \(A B C D\) is a parallelogram. (c) Let \(A B C D\) be a quadrilateral in which \(\angle A=\angle C\) and \(\angle B=\angle D\). Prove that \(A B \| D C\) and that \(B C \| A D,\) and hence that \(A B C D\) is a parallelogram.

Short Answer

Expert verified
In all three parts, given conditions imply that quadrilateral \(ABCD\) is a parallelogram by properties of parallel lines and equal sides or angles.

Step by step solution

01

Understanding Part (a)

In part (a), we are given that in quadrilateral \(ABCD\), \(AB \parallel DC\) and \(\underline{AB} = \underline{DC}\). We need to prove that \(BC \parallel AD\) to show that \(ABCD\) is a parallelogram.
02

Proving Part (a) Using Properties of Parallelograms

Given \(AB \parallel DC\) and \(\underline{AB} = \underline{DC}\), we can start by noting the properties of trapezoids since \(AB\) and \(DC\) are parallel. If the pair of parallel lines in a trapezoid are equal in length, it becomes an isosceles trapezoid. In an isosceles trapezoid, both the non-parallel sides are equal in length and parallel as well. Thus, \(BC \parallel AD\), making \(ABCD\) a parallelogram.
03

Understanding Part (b)

In part (b), we are given that \(\underline{AB} = \underline{DC}\) and \(\underline{BC} = \underline{AD}\). We need to show that \(AB \parallel DC\) and hence \(ABCD\) is a parallelogram.
04

Proving Part (b) Using Opposite Sides

If both pairs of opposite sides of a quadrilateral are equal, i.e., \(\underline{AB} = \underline{DC}\) and \(\underline{BC} = \underline{AD}\), then by the properties of parallelograms, the opposite sides must be parallel. Therefore, \(AB \parallel DC\) and \(BC \parallel AD\), concluding that \(ABCD\) is a parallelogram.
05

Understanding Part (c)

In part (c), we are given that \(\angle A = \angle C\) and \(\angle B = \angle D\). We need to prove \(AB \parallel DC\) and \(BC \parallel AD\), leading to \(ABCD\) being a parallelogram.
06

Proving Part (c) Using Angle Properties

For a quadrilateral, if alternate interior angles are equal (\(\angle A = \angle C\)) and (\(\angle B = \angle D\)), this indicates that line \(AB\) is parallel to line \(DC\) and line \(BC\) is parallel to line \(AD\). This configuration is consistent with a parallelogram, confirming that \(ABCD\) is a parallelogram.

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Key Concepts

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

Quadrilaterals
Quadrilaterals are four-sided polygons with various properties that make them interesting in geometry. They are defined by four vertices, four edges, and four angles. Understanding quadrilaterals is pivotal in solving many geometric problems, as they serve as the basis for more complex shapes.

In general, quadrilaterals can take many forms, such as squares, rectangles, rhombuses, and trapezoids. Differentiating between these forms involves looking at properties like side lengths and angles.
  • A square has equal sides and four right angles.
  • A rectangle also has four right angles, but doesn鈥檛 necessarily have equal side lengths.
  • A rhombus has equal side lengths but angles that are not necessarily right angles.
  • A trapezoid has at least one pair of parallel sides.

Mastering these differences is crucial for identifying and proving the characteristics of quadrilaterals like parallelograms.
Parallel Lines
Parallel lines are a fundamental concept in geometry and essential for understanding quadrilaterals. When two lines are parallel, they run side by side and will never meet, no matter how long they are extended. This concept is crucial when determining the types of quadrilaterals.

In quadrilaterals, having parallel lines often assists in establishing the shape as a parallelogram.
  • For example, a parallelogram has two pairs of parallel sides.
  • These parallel sides mean the opposite angles of a parallelogram are equal.
  • Parallel lines are often denoted in diagrams with arrows pointing in the same direction.

Understanding how parallel lines define the structure of quadrilaterals is important in geometric proofs and helps determine specific properties of these shapes.
Geometric Proofs
Geometric proofs are logical arguments used to prove the properties of geometric figures. They rely on previously established theorems and postulates. Proofs often involve a series of statements and reasons that build on one another to reach a conclusion.

For quadrilaterals, geometric proofs are used to establish characteristics like side lengths, angle measures, and parallelism.
  • Each statement in a proof must be justified with a theorem or property.
  • In the case of parallelograms, properties such as opposite sides being equal and opposite angles being equal are frequently used.
  • Geometric proofs require attention to detail and a clear understanding of the relationships between various elements of the figure.

By practicing proofs, students develop a deeper understanding of geometric concepts and the logical methods needed to solve geometric problems.
Properties of Parallelograms
Parallelograms are a special type of quadrilateral with specific properties that help distinguish them from other quadrilaterals. Recognizing these properties is essential in geometric problem-solving.

Some of the key properties of parallelograms include:
  • Opposite sides are parallel and equal in length.
  • Opposite angles are equal.
  • The diagonals bisect each other.
  • The sum of the angles in a parallelogram is always 360 degrees.

Understanding these properties helps in proving whether a given quadrilateral is a parallelogram. For instance, if you can show that both pairs of opposite sides are equal, then the quadrilateral must be a parallelogram. This forms the basis for solving problems like the ones in the exercise above.

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Most popular questions from this chapter

Suppose that the line \(X A Y\) is tangent to the circumcircle of \(\triangle A B C\) at the point \(A,\) and that \(X\) and \(C\) lie on opposite sides of the line \(A B .\) Prove that \(\angle X A B=\angle A C B\)

Given \(\triangle A B C,\) let \(L\) be the midpoint of the side \(\underline{B C}\). The line \(A L\) is called a median of \(\triangle A B C\). (It is not at all obvious, but if we imagine the triangle as a lamina, having a uniform thickness, then \(\triangle A B C\) would exactly balance if placed on a knife- edge running along the line \(A L\).) Let \(M\) be the midpoint of the side \(\underline{C A}\), so that \(B M\) is another median of \(\triangle A B C .\) Let \(G\) be the point where \(A L\) and \(B M\) meet. (a)(i) Prove that \(\triangle A B L\) and \(\triangle A C L\) have equal area. Conclude that \(\triangle A B G\) and \(\triangle A C G\) have equal area. (ii) Prove that \(\triangle B C M\) and \(\triangle B A M\) have equal area. Conclude that \(\triangle B C G\) and \(\triangle B A G\) have equal area. (b) Let \(N\) be the midpoint of \(\underline{A B}\). Prove that \(C G\) and \(G N\) are the same straight line (i.e. that \(\angle C G N\) is a straight angle). Hence conclude that the three medians of any triangle always meet in a point \(G\).

A regular \(2 n\) -gon \(A B C D E \cdots\) is inscribed in a circle of radius \(r\). The \(2 n\) radii \(O A, O B, \ldots\) joining the centre \(O\) to the \(2 n\) vertices cut the circle into \(2 n\) sectors, each with angle \(\frac{\pi}{n}\) (Figure 6 ). These \(2 n\) sectors can be re-arranged to form an "almost rectangle", by orienting them alternately to point "up" and "down". In what sense does this "almost rectangle" have "height \(=r "\) and "width \(=\pi r " ?\)

(a) Given points \(A, B,\) with \(\underline{A B}=2 c,\) and a positive real number \(a\). Find the locus of all points \(X\) such that \(|\underline{A X}-\underline{B X}|=2 a\). (b) Given a point \(F\) and a line \(m,\) find the locus of all points \(X\) such that the ratio distance from \(X\) to the point \(F:\) distance from \(X\) to the line \(m\) is a constant \(e>1\) (c) Prove that parts (a) and (b) give different ways of specifying the same curve, or locus.

Given any triangle \(\triangle A B C,\) extend \(\underline{B C}\) beyond \(C\) to a point \(X\). Then the exterior angle $$\angle X C A=\angle A+\angle B$$ ("In any triangle, each exterior angle is equal to the sum of the two interior opposite angles.") Another important consequence is the result which underpins the sequence of "circle theorems".
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