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Q17E

Expert-verifiedFound in: Page 470

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 the**

**constructible 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 ${\text{CQ}}$** **is**

**perpendicular to ** ${\text{L}}$**.]**

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

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.**

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

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

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 $\text{L}$.

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

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

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

Figure 6

$\text{DQ}$, 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.

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