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Chapter 5: Problem 157

Let \(A B C D\) be a parallelogram. (i) Prove that \(\triangle A B C \equiv \triangle C D A,\) so that each triangle has area exactly half of area \((A B C D)\) (ii) Conclude that opposite sides of \(A B C D\) are equal in pairs and that opposite angles are equal in pairs. (iii) Let \(A C\) and \(B D\) meet at \(X\). Prove that \(X\) is the midpoint of both \(\underline{A C}\) and \(\underline{B D} .\)

Short Answer

Expert verified
\(\triangle ABC \equiv \triangle CDA\); opposite sides and angles are equal; diagonals bisect each other at midpoints.

Step by step solution

01

Identify Triangle Correspondence

Observe that in parallelogram \(ABCD\), triangles \(\triangle ABC\) and \(\triangle CDA\) share side \(AC\). Also, \(AB\parallel CD\) and \(AD\parallel BC\) by the definition of a parallelogram. The angles \(\angle BAC = \angle DCA\) and \(\angle ACB = \angle CAD\) because they are alternate interior angles with parallel lines cut by transversal \(AC\).
02

Prove Triangle Congruence

Using ASA (Angle-Side-Angle) congruence theorem, \(\triangle ABC \equiv \triangle CDA\) because \(AB = CD\), \(\angle BAC = \angle DCA\), \(\angle ACB = \angle CAD\).
03

Area Relation in Parallelogram

Since \(\triangle ABC \equiv \triangle CDA\), they have the same area. Hence, each triangle's area is exactly half of the area of parallelogram \(ABCD\). Therefore, \(\text{Area}(\triangle ABC) = \frac{1}{2} \times \text{Area}(ABCD)\).
04

Opposite Sides and Angles Equality

By the properties of parallelogram \(ABCD\), opposite sides are equal: \(AB = CD\) and \(AD = BC\). Also, since the alternate interior angles are equal, opposite angles are equal: \(\angle B = \angle D\) and \(\angle A = \angle C\).
05

Intersection of Diagonals at Midpoint

The diagonals of parallelogram \(ABCD\) bisect each other. Therefore, point \(X\) is the midpoint of both diagonals \(AC\) and \(BD\). This can be shown by stating that since opposite sides are equal and parallel, and using the midpoint theorem, diagonals bisect each other, making \(X\) the midpoint of both \(AC\) and \(BD\).

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Key Concepts

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

Parallelogram Properties
A parallelogram is a special type of quadrilateral with unique characteristics. If you look at any parallelogram, you'll notice that its opposite sides are parallel. This trait also means the opposite sides are equal in length. As a result, in parallelogram ABCD:
  • AB is equal to CD.
  • AD is equal to BC.
Moreover, opposite angles are equal, which is essential when considering the shape's symmetry. This means:
  • Angle A is equal to Angle C.
  • Angle B is equal to Angle D.
These properties not only make calculations involving parallelograms simpler but play a crucial role in solving complex geometric problems involving these shapes.
Triangle Congruence
The concept of triangle congruence is paramount in geometry because it allows us to establish the equivalence of two triangles' shapes and sizes. In a parallelogram, such as ABCD, we use the Angle-Side-Angle (ASA) criterion to prove congruence. To apply ASA, we need:
  • Two angles and the included side to be congruent to another triangle's corresponding angles and side.
For triangles ABC and CDA:
  • Side AC is shared, making it equivalent itself.
  • Due to parallelism and transversal line, alternate interior angles BAC and DCA are equal, as well as ACB and CAD.
Hence, these conditions let us safely declare that triangles ABC and CDA are congruent by ASA.
Area of Parallelogram
Determining the area of a parallelogram involves breaking it down into simpler components. With triangles ABC and CDA congruent, each has an area equal to half of the parallelogram's total area. Here's why:
  • These triangles are mirror images along the shared diagonal AC within the shape.
  • Their congruence ensures their areas are equal.
So, the area of parallelogram ABCD can be expressed as:\[\text{Area}(\triangle ABC) = \frac{1}{2} \times \text{Area}(ABCD)\]This understanding is vital because it helps when dealing with more complex calculations involving parallelograms, like determining side lengths or angles when given the area.
Diagonal Intersection Theorem
The intersection of diagonals in a parallelogram leads us to the midpoint theorem. For a parallelogram ABCD with diagonals AC and BD intersecting at point X, this theorem holds firm. It states:
  • The diagonals bisect each other at the point of intersection.
This means:
  • X is the midpoint of diagonal AC.
  • X is also the midpoint of diagonal BD.
This bisection confirms that each half of the diagonals is equal, further substantiating the congruent triangles formed within the parallelogram. Understanding this theorem is fundamental when diving deeper into parallelogram-related problems, as it imparts symmetrical insight about the shape's structure.

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Most popular questions from this chapter

Let \(A B C D\) be a trapezium with \(A B \| D C,\) in which \(A B\) has length \(a\) and \(D C\) has length \(b\). (a) Let \(M\) be the midpoint of \(\underline{A D}\) and let \(N\) be the midpoint of \(\underline{B C}\). Prove that \(M N \| A B\) and find the length of \(\underline{M N}\). (b) If the perpendicular distance between \(A B\) and \(D C\) is \(d,\) find the area of the trapezium \(A B C D\).

(The Midpoint Theorem) Given any triangle \(\triangle A B C,\) let \(M\) be the midpoint of the side \(\underline{A C},\) and let \(N\) be the midpoint of the side \(\underline{A B}\). Draw in \(M N\) and extend it beyond \(N\) to a point \(M^{\prime}\) such that \(\underline{M N}=\underline{N M^{\prime}}\). (a) Prove that \(\triangle A N M \equiv \Delta B N M^{\prime}\). (b) Conclude that \(\underline{B M^{\prime}}=\underline{C M}\) and that \(B M^{\prime} \| C M\). (c) Conclude that \(M M^{\prime} B C\) is a parallelogram, so that \(\underline{C B}=\underline{M M^{\prime}}\). Hence \(\underline{M N}\) is parallel to \(\underline{C B}\) and half its length.

(a) Construct a right angled triangle that explains the standard formula for the distance from \(P=(a, b)\) to \(Q=(d, e)\). (b) Use part (a) to derive the standard formula for the distance from \(P=\) \((a, b, c)\) to \(Q=(d, e, f)\)

Given three points \(A, B, C,\) show how to construct \(-\) without measuring \(-\) a point \(D\) such that the segments \(\underline{A B}\) and \(\underline{C D}\) are equal (in length).

(a) Five vertices \(A, B, C, D, E\) are arranged in cyclic order. However instead of joining each vertex to its two immediate neighbours to form a convex pentagon, we join each vertex to the next but one vertex to form a pentagonal star, or pentagram \(A C E B D .\) Calculate the sum of the five "angles" in any such pentagonal star. (b) There are two types of 7 -gonal stars. Calculate the sum of the angles at the seven vertices for each type. (c) Try to extend the previous two results (and the proofs) to arbitrary \(n\) -gonal stars.
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