Let be a constructible point and a constructible line. Prove that the linethrough perpendicular to is constructible. [Hint: The case when is on was done in Example 1. If is not on and is a constructible point on , the circle with center and radius is constructible and meets at the
constructible points and . The circles with center , radius and center
, radius intersect at constructive points and . Show that line is
perpendicular to .]
The line through a constructible point and perpendicular to 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 and the line , which are both constructible.
Assume that is a constructible point on the line , and that is also a constructible point on the line . Draw a circle with radius that meets the line at points and using the compass at constructible point (as in figure ).
Make two circles, one with the center at and the radius , and the other with the center at and the radius . Assume that the circles overlap at A and B (as in figure ).
Let consider triangles and (as in figure ).
and are both equivalent to on both sides. The triangles' sides and are equal, while the side is shared by both. As a result of the side-side-side congruence rule, the triangles and are congruent.
As a result, the angles and are complementary angles. As a result, the line is parallel to the line .
Let's assume that point is not on the constructible line . Consider the point on the line L that is constructible.
Draw a circle of radius that meets the line at the points and by placing the compass point at (as in figure ).
Make two circles, one with a center at and a radius of , and the other with a center at and a radius of . Consider the intersection of the two circles at points and (as in figure ).
Take a closer look at the and triangles (as in figure ).
, being the radius of the circle with the center at is equal to the side .
Both triangles have the same side , and the side equals (since the circle of the circle bisects the chord).
As a result of the side-side-side congruence rule, the triangles and are congruent.
As both the angles and are equivalent and complementary, so the angle is measured at 90 degrees. As a result, the line is parallel to the line .
The claim's proof is now complete.
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