MEDAL & FAN ANYBODY ? Using a compass, constructing a line parallel to a given line requires: a. copying an angle using a point NOT on the given line as its vertex. b. drawing a ray from the given line to a point NOT on the line. c. bisecting the given line. d. drawing three collinear points from the given line.
Task. Construct the line parallel to a given line ℓ and passing through a given point P which is not on ℓ. . ℓ P Solution. 1. Draw a circle c1 with center P and intersecting ℓ at two points, one of which is A. . . . ℓ P A 2. Draw a second circle c2 with center A and the same radius r as c1. This circle also intersects ℓ at two points, one of which is B. . . . ℓ P A B 3. Draw a third circle c3 with center B and radius r. Let C be the intersection point of c3 (drawn below in red) with c1 (drawn below in green) which lies on the same side of ℓ as P does. The line PC (drawn below in blue) is the required parallel to ℓ. . . . ℓ P A B C Note 1. The construction is based on the fact that the quadrilateral PABC is a parallelogram. In fact, PABC is a rhombus. The reasoning is as follows: The green circle shows that PC⎯⎯⎯⎯⎯ and PA⎯⎯⎯⎯⎯ are congruent. The black circle shows that PA⎯⎯⎯⎯⎯ and AB⎯⎯⎯⎯⎯ are congruent. The red circle shows that AB⎯⎯⎯⎯⎯ and BC⎯⎯⎯⎯⎯ are congruent. Since PABC is a quadrilateral with all sides congruent, it is a rhombus (and therefore a parallelogram). Note 2. It is clear that the construction only needs the compass, not a straightedge: In determining the point C, the straightedge is totally superfluous, and the points P and C determine the desired line (which thus is not necessary to actually draw!). It may be proved that all constructions with compass and straightedge are possible using only the compass. Note 3. Another construction of the parallel uses the fact that the endpoints of two congruent chords (red) in a circle determine two parallel chords
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@iamawesome1 that was wrong...
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