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Chapter 5: Problem 166

(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.

Short Answer

Expert verified
Triangle \( \triangle ANM \equiv \triangle BNM' \), \( \overline{BM'} = \overline{CM} \) and \( BM' \parallel CM \). Quadrilateral \( MM'BC \) is a parallelogram, so \( \overline{CB} = \overline{MM'} \) and \( \overline{MN} \parallel \overline{CB} \).

Step by step solution

01

Analyze Points and Structure

In triangle \( \triangle ABC \), identify points \( M \) and \( N \) as midpoints of \( \overline{AC} \) and \( \overline{AB} \), respectively. Extend line \( \overline{MN} \) to a point \( M' \) such that \( \overline{MN} = \overline{NM'} \). This defines \( \overline{AN} = \overline{NB} \) and \( \overline{AM} = \overline{MC} \) as midpoints by triangle properties.
02

Prove Congruence of Triangles

To prove \( \triangle ANM \equiv \triangle BNM' \), we use the Side-Angle-Side (SAS) theorem for congruence. Note that \( \overline{AN} = \overline{NB} \), \( \angle ANM = \angle BNM' \) as corresponding angles, and \( \overline{NM} = \overline{NM'} \) by construction.
03

Deduce Equal Lengths and Parallelism

From the congruence \( \triangle ANM \equiv \triangle BNM' \), it follows that \( \overline{AM} = \overline{BM'} \). Since \( \overline{NM} \parallel \overline{NM'} \), and equal measurements make the corresponding sides \( \overline{BM'} \parallel \overline{CM} \) and equal in length.
04

Establish Parallelogram Properties

Since \( \overline{BM'} \parallel \overline{MC} \) and \( \overline{NM} = \overline{NM'} \), quadrilateral \( MM'BC \) is a parallelogram. That makes \( \overline{CB} = \overline{MM'} \), and \( \overline{MN} \parallel \overline{CB} \), maintaining that \( \overline{MN} \) is half the length of \( \overline{CB} \).

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Key Concepts

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

Triangle Congruence
In geometry, triangle congruence is a critical concept. Two triangles are congruent if they have exactly the same size and shape. This means all corresponding sides and angles of the triangles are equal. When proving triangle congruence, there are several postulates to consider, such as Side-Angle-Side (SAS), which was utilized in solving the original exercise. In our case, the triangles \( \triangle ANM \) and \( \triangle BNM' \) were shown to be congruent by SAS. This was because:
  • \( \overline{AN} = \overline{NB} \) since both N is the midpoint of \( \overline{AB} \)

  • \( \angle ANM = \angle BNM' \) as they are vertically opposite angles.

  • \( \overline{NM} = \overline{NM'} \) by construction.
Proving congruence in two triangles like this helps in concluding further properties, such as determining parallel lines and equal segments.
Parallelogram Properties
Parallelograms are special quadrilaterals where opposite sides are parallel and equal in length. Recognizing parallelograms often helps solve complex geometric problems. In the given exercise, understanding parallelogram properties was key to establishing that \( MM'M'C \) is a parallelogram.
  • The construction started by showing that \( \overline{BM'} \parallel \overline{MC} \) and are equal due to congruent triangles \( \triangle ANM \equiv \triangle BNM' \).

  • Since \( \overline{MN} \equiv \overline{NM'} \), and both are parallel to their counterparts, \( MM'M'C \) is confirmed as a parallelogram.
This property then allowed the insight that \( \overline{CB} = \overline{MM'} \). This understanding of parallelogram properties leads to deducing other important features such as parallelism and equal lengths in further calculations.
Parallel Lines
Parallel lines are fundamental in establishing many geometric theorems and properties, including the Midpoint Theorem applied in the exercise. Two lines are parallel if they are always the same distance apart and never meet. In the context of the given problem:
  • It was found that \( \overline{BM'} \parallel \overline{MC} \) due to the congruent relation of \( \triangle ANM \equiv \triangle BNM' \).

  • The identification of \( MM'M'C \) as a parallelogram allowed us to state that \( \overline{MN} \parallel \overline{CB} \), and importantly, \( \overline{MN} \) is half the length of \( \overline{CB} \).
This understanding that segments linked by the midpoints maintain parallel conditions helps in solving further geometric problems efficiently and confidently.

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Most popular questions from this chapter

Given a circle with centre \(O,\) let \(Q\) be a point outside the circle, and let \(Q P, Q P^{\prime}\) be the two tangents from \(Q,\) touching the circle at \(P\) and at \(P^{\prime} .\) Prove that \(Q P=Q P^{\prime},\) and that the line \(O Q\) bisects the angle \(\angle P Q P^{\prime}\)

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) 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.

You are given two lines \(m\) and \(n\) crossing at the point \(B\). (a) If \(A\) lies on \(m\) and \(C\) lies on \(n,\) prove that each point \(X\) on the bisector of angle \(\angle A B C\) is equidistant from \(m\) and from \(n\). (b) If \(X\) is equidistant from \(m\) and from \(n,\) prove that \(X\) must lie on one of the bisectors of the two angles at \(B\).

(a) In the quadrilateral \(A B C D\) the two diagonals \(\underline{A C}\) and \(\underline{B D}\) cross at \(X\). Suppose \(\underline{A B}=\underline{B C}, \angle B A C=60^{\circ}, \angle D A C=40^{\circ}, \angle B X C=100^{\circ}\) (i) Calculate (exactly) \(\angle A D B\) and \(\angle C B D\). (ii) Calculate \(\angle B D C\) and \(\angle A C D\). (b) In the quadrilateral \(A B C D\) the two diagonals \(\underline{A C}\) and \(\underline{B D}\) cross at \(X\). Suppose \(\underline{A B}=\underline{B C}, \angle B A C=70^{\circ}, \angle D A C=40^{\circ}, \angle B X C=100^{\circ}\) (i) Calculate (exactly) the size of \(\angle B D C+\angle A C D\) (ii) Explain how we can be sure that \(\angle B D C\) and \(\angle A C D\) are uniquely determined, even though we cannot calculate them immediately. \(\triangle\)
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