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Q17E

Expert-verified
Found in: Page 470

### Abstract Algebra: An Introduction

Book edition 3rd
Author(s) Thomas W Hungerford, David Leep
Pages 608 pages
ISBN 9781111569624

# Let $\text{C}$ be a constructible point and ${\text{L}}$ a constructible line. Prove that the linethrough ${\text{C}}$ perpendicular to ${\text{L}}$ is constructible. [Hint: The case when ${\text{C}}$ is on was done in Example 1. If is not on ${\text{L}}$ and ${\text{D}}$ is a constructible point on ${\text{L}}$, the circle with center ${\text{C}}$ and radius ${\text{CD}}$ is constructible and meets ${\text{L}}$ at theconstructible points ${\text{D}}$ and ${\text{E}}$. The circles with center ${\text{D}}$, radius ${\text{CD}}$ and center ${\text{E}}$, radius ${\text{CE}}$ intersect at constructive points ${\text{C}}$ and $\text{Q}$. Show that line ${\text{CQ}}$ isperpendicular to ${\text{L}}$.]

The line through a constructible point $\text{C}$ and perpendicular to $\text{L}$ is constructible.

See the step by step solution

## Step 1: Introduction

To coordinate the plane, consider any two points and treat the distance between them as a unit length. Any point in the coordinate plane that may be built with straightedge and compass constructions is referred to as constructible. If a line goes through at least two constructible points, it is said to be constructible.

## Step 2: Assuming the claim to be true

Let’s suppose that a line through a constructible point and perpendicular to a constructible line is constructible. Now, consider the point $\text{C}$ and the line $\text{L}$ , which are both constructible.

Assume that $\text{C}$ is a constructible point on the line $\text{L}$, and that $\text{D}$ is also a constructible point on the line $\text{L}$. Draw a circle with radius $\text{CD}$ that meets the line at points $\text{D}$ and $\text{E}$ using the compass at constructible point $\text{C}$ (as in figure ).

Figure 1

## Step 3: Constructing circles

Make two circles, one with the center at $\text{E}$ and the radius $\text{ED}$, and the other with the center at $\text{D}$ and the radius $\text{ED}$. Assume that the circles overlap at A and B (as in figure $\text{2}$ ).

Figure 2

## Step 4: Considering triangles

Let consider triangles $\text{AEC}$ and $\text{ACD}$ (as in figure$\text{3}$ ).

Figure 3

$\text{AE}$ and $\text{AD}$ are both equivalent to $\text{ED}$ on both sides. The triangles' sides $\text{EC}$ and $\text{CD}$ are equal, while the side $\text{AC}$ is shared by both. As a result of the side-side-side congruence rule, the triangles $\text{AEC}$ and $\text{ACD}$ are congruent.

As a result, the angles $\text{AEC}$ and $\text{ACD}$ are complementary angles. As a result, the line $\text{AC}$ is parallel to the line $\text{L}$.

## Step 5: Assuming C  is not on line L

Let's assume that point $\text{C}$ is not on the constructible line $\text{L}$. Consider the point $\text{D}$ on the line L that is constructible.

Draw a circle of radius $\text{CD}$ that meets the line $\text{L}$ at the points $\text{C}$ and $\text{D}$ by placing the compass point at $\text{C}$ (as in figure $\text{4}$ ).

Figure 4

## Step 6: Constructing circles and considering its intersection

Make two circles, one with a center at $\text{D}$ and a radius of ,$\text{CD}$ and the other with a center at $\text{E}$ and a radius of $\text{CE}$. Consider the intersection of the two circles at points $\text{C}$ and $\text{Q}$ (as in figure $\text{5}$ ).

Figure 5

## Step 7: Considering the triangles

Take a closer look at the $\text{DOQ}$ and $\text{DOC}$ triangles (as in figure$\text{6}$ ).

Figure 6

## Step 8: Claim conclusion

$\text{DQ}$, being the radius of the circle with the center at $\text{D}$ is equal to the side $\text{CD}$$\text{CD}$.

Both triangles have the same side $\text{DO}$ , and the side $\text{CO}$equals $\text{OQ}$ (since the circle of the circle bisects the chord).

As a result of the side-side-side congruence rule, the triangles $\text{DOC}$ and $\text{DOQ}$ are congruent.

As both the angles $\text{DOC}$ and $\text{DOQ}$ are equivalent and complementary, so the angle $\text{DOC}$ is measured at 90 degrees. As a result, the line $\text{CQ}$ is parallel to the line $\text{L}$.

The claim's proof is now complete.

Hence,Proved.