UNIT 3.6 CONSTRUCTING PARALLEL AND

PERPENDICULAR LINES

Remember -- use your compass and straight edge only!

Given: Point P is off a given lineTask: Construct a line through P parallel to the given line

Directions:

1. With your straightedge, draw a transversal through point P. This is simply a straight line which runs through P and intersects the given line.2. Using your knowledge of the construction COPY AN ANGLE, construct a copy of the angle formed by the transversal and the given line such that the copy is located UP at point P. The vertex of your copied angle will be point P.3. When the copy of the angle is complete, you will have two parallel lines.

This new line is parallel to the given line.

Explanation of construction: Since we used the construction to copy an angle, we now have two angles of equal measure in our diagram. In relation to parallel lines, these two equal angles are positioned in such a manner that they are called corresponding angles. A theorem relating to parallel lines tells us that if two lines are cut by a transversal and the corresponding angles are congruent (equal), then the lines are parallel.

Perpendicular - lines (or segments) which meet to form right angles.

Given:  Point P is off a given lineTask:  Construct a line through P perpendicular to the given line.

1. Place your compass point on P and sweep an arc of any size that crosses the line twice. 2. Place the compass point where the arc crossed the line on one side and make an arc ON THE OPPOSITE SIDE OF THE LINE. 3. Without changing the span on the compass, place the compass point where the arc crossed the line on the OTHER side and make another arc. Your two new arcs should be crossing on the opposite side of the line.4. With your straightedge, connect the intersection of the two new arcs to point P.

This new line is perpendicular to the given line.

Explanation of construction: To understand the explanation, some additional labeling will be needed. Label the point where the arc crosses the line as points C and D. Label the intersection of the new arcs on the opposite side as point E. Draw segments , , , and . By the construction, PC = PD and EC = ED. Now, remember a locus theorem: The locus of points equidistant from two points (C and D), is the perpendicular bisector of the line segment determined by the two points. Hence, is the perpendicular bisector of . The fact that we created a bisector, as well as a perpendicular, is actually MORE than we needed - we only needed to create a perpendicular.

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