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In this lesson, you will study definitions for the following objects: complementary and supplementary angles, angle bisectors, and perpendicular bisector of a line segment. You will also construct these objects using a straightedge and compass.
Existence and Uniqueness of Parallel Lines Let L be any line and P be a point not on L. Then there is
only one line containing P parallel to L. There is also an analogue to the theorem above for perpendicular lines.
Although we may have implicitly referred to supplementary and complementary angles in previous sections, let us define them here for completeness.
If the sum of the measures of two angles is 180°, then the angles are called supplementary, and each angle is called the supplement of the other.
If the sum of the measures of two angles is 90°, then the angles are called complementary and each angle is called the complement of the other.
In this section, we will be dealing with angles formed by rays, which are parts of a line that start at a point and extend infinitely in the other direction. Below, 
BAC is formed by the rays 
and 
. We’ll start off with the simplest construction— angle supplements.
Suppose we are given an angle BAC as above where mBAC < 180°. We can construct the supplement to BAC with the same vertex A using a straightedge.
Step 1: Extend the line containing the ray in the opposite direction using your straightedge.
Step 2: Choose a point D on this new ray.
The angle DAC that we have just constructed is the supplementary angle to 
BAC because 
.
The angle bisector of BAC is the ray 
, such that BAD 
DAC. In other words, an angle bisector divides an angle into two congruent angles.
Suppose we are given A.
Step 1: Using A as a center we draw an arc of a circle intersecting the two sides of the angle. Label these points B and C. The segments 
and 
are congruent (these are both radii of the circle).
Step 2: Draw the circle with center B and radius r = BC.
Step 3: Draw the circle with center C and radius r = BC.
Step 4: These two circles will intersect at exactly two points. Choose the point on the other side of the segment BC and label this point P. Draw the ray originating at A through P. This ray is the angle bisector of BAC.
In steps 2 and 3, we could have used any radius greater than 
. This ensures that the circles will intersect.
Let’s try an example. When constructing an angle on a given side of a ray congruent to a given A,
what is the minimum number of arcs (or circles) necessary to construct the angle?
The correct answer is 3. We need to draw the first arc on the given angle A to create a triangle ABC with congruent sides r = AB = AC and a third side BC = s.
We draw a circle of radius r with center D that intersects the given ray at a point E. This gives us one side of a triangle that will eventually be congruent to BAC.
Next, we draw a circle of radius s and center E. This intersects this first circle in two points, but we choose the one on the given side. The ray drawn through this point (F) creates the desired congruent triangle (by SSS), and thus the corresponding angles A and D we have constructed are also congruent.
Suppose we are given a line L and a point P that is not on L. Let Q and R be any two points on L.
Step 1: To construct a parallel line to point P, first draw the line 
.
Step 2: We can draw angle QPS congruent to angle PQR on the opposite side of from R. These two angles, QPS and PQR, are alternate interior angles.
Because these angles are congruent, the line 
is parallel to the line L.
The perpendicular bisector of a line segment is the line perpendicular to the segment at its midpoint.
Suppose we are given a line segment 
.
Step 1 : To construct a perpendicular bisector, first draw a circle with center B and radius r = AB.
Step 2: Draw a circle with center at A and radius r = AB.
Step 3: The two circles intersect at two points on opposite sides of . Label these points P and Q. Draw 
.
PQ is equidistant from A and B, and is, therefore, the perpendicular bisector of .
The process of constructing a line perpendicular to a given line containing an external point is quite similar to the construction we have just completed. Basically, all we need to do is find the right segment to bisect. So let’s just do this one step.
Again suppose we are given a line L and a point P that is not on L. Draw a circle with center P and with large enough radius so that it intersects L in two points, Q and R . If we construct the perpendicular bisector of QR, the line passes through P, because P is equidistant from R and S. So by the previous construction, the perpendicular bisector of QR contains P and is perpendicular to L.
If you recall, we gave the definition of an angle complement at the very beginning of the section. This construction follows from work we’ve already practiced.
The construction of an angle complementary to a given A sharing the same vertex, A, is a corollary of the construction of a perpendicular bisector.
If we extend one ray of angle A to a line and create a segment BC, such that A is the midpoint of 
, then the perpendicular bisector of will give us a ray, AD, which creates the angle complement to angle A.
Suppose we are given a line segment 
that we wish to subdivide into n congruent subsegments. The figure below shows the case for n = 5.
Step 1: Draw any ray originating at the point A, but not containing B. Name this ray 
Step 2: Draw an arc of a circle centered at A with positive radius r and intersecting L at a point P1.
Step 3: Draw an arc of a circle centered at P1 with the same radius r and intersecting L at a point P2.
Step 4: Repeat step 3 until you reach Pn . At this point you’ve drawn n congruent segments on L:P1, P1P2, P2P3, . . . Pn – 1Pn.
Step 5: Using your straightedge, connect Pn to B.
Step 6: Through the other points P1, P2, . . .Pn – 1, draw lines parallel to Pn B, intersecting at points Q1, Q2,…Qn–1, respectively.
Because the parallel lines intercept congruent segments on L, they intersect congruent segments on .