Challenge 09: Construct the Midpoint
- Author:
- Gerry Stahl
Circles are very useful for constructing figures with dependencies. Because all radii of a circle are the same length, you can make the length of one segment be dependent on the length of another segment by constructing both segments to be radii of the same circle. This is what Euclid did to construct an equilateral triangle. Circles can also be used for constructing the midpoint of a segment or perpendicular lines or parallel lines, for instance.
As you already saw, the construction process for an equilateral triangle creates a number of interesting relationships among different points and segments. In this challenge, points A, B and C form an equilateral triangle. Segment AB crosses segment CD at the exact midpoint of CD and the angles between these two segments are all right angles (90 degrees). We say that AB is the “perpendicular bisector” of CD—meaning that AB cuts CD at its midpoint, evenly into two equal-length segments, and that AB is perpendicular (meaning, at a right angle) to CD.
To find the midpoint of a segment AB, construct circles of radius AB centered on A and on B. Construct points C and D at the intersections of the circles. Segment CD intersects segment AB at the midpoint of AB.