Geometric Constructions
A free Geometry lesson from the “Geometry Foundations” unit, with a worked example and practice problems including step-by-step solutions.
A construction produces a figure using permitted tools, then explains why the figure has the required property. To construct the perpendicular bisector of a segment, keep the compass radius fixed and greater than half the segment length. Equal-radius arcs from the endpoints intersect in two points. Their connecting line is the perpendicular bisector; its correctness follows from congruent triangles, not from how the drawing looks.
What you'll learn
- Construct a perpendicular bisector of a segment using compass and straightedge
- Construct an angle bisector
- Construct a copy of a segment and a copy of an angle
- Recognize that construction tools force exactness rather than approximation
Worked example
Problem. Construct the perpendicular bisector of segment AB using a compass and an unmarked straightedge, and justify the result.
- Draw a circle or arcs centered at A with radius greater than AB/2. Without changing the radius, repeat with center B. Label the intersections P and Q.
- Draw line PQ and let M be its intersection with AB. PA = PB and QA = QB because all four are compass radii; PQ is shared.
- Triangles APQ and BPQ are congruent by SSS. Therefore angle APM equals angle BPM (M lies on the ray from P through Q).
- Triangles APM and BPM are congruent by SAS: PA = PB, PM is shared, and the included angles at P are equal. Thus AM = BM and angles AMP and BMP are equal.
- A, M, and B are collinear, so the equal adjacent angles at M sum to 180 degrees. Each is 90 degrees. Hence PQ is the perpendicular bisector.
Answer: The line through the two equal-radius arc intersections bisects AB at a right angle.
Lesson preview: this page shows 5 practice items. The interactive activity contains 28 practice items and 15 quiz items (multiple choice and typed response). Counts include repeated prompts and variants.
Practice problems
1. Which two tools are allowed in classical geometric constructions?
Choices: Ruler and protractor · Compass and straightedge · Compass and protractor · Straightedge and graph paper
Show solution
- Constructions use a compass for circles and a straightedge for lines.
- Rulers and protractors measure — they would not produce exact results.
Answer: Compass and straightedge
2. A straightedge is different from a ruler because it has no:
Choices: Markings or units · Color · Edge · Length
Show solution
- A straightedge draws lines; it does not measure them.
Answer: Markings or units
3. To copy a segment, you set the compass to the segment's length and then:
Choices: Draw a circle of that radius from the new endpoint · Use a protractor · Trace it freehand · Estimate with a ruler
Show solution
- The radius equals the segment length; the arc marks the new endpoint.
Answer: Draw a circle of that radius from the new endpoint
4. To bisect segment AB with compass and straightedge, the compass radius must be:
Choices: More than half of AB · Exactly half of AB · The full length of AB · Less than half of AB
Show solution
- Smaller arcs would not intersect.
- Equal arcs > half AB produce two intersections on the perpendicular bisector.
Answer: More than half of AB
5. After drawing the two intersecting arcs, the next step in bisecting a segment is to:
Choices: Draw a circle through both endpoints · Connect the two arc intersections with a straightedge · Erase the arcs · Measure the angle
Show solution
- The line through both intersections is perpendicular to the segment AND passes through its midpoint.
Answer: Connect the two arc intersections with a straightedge
Explain and apply
Complete these tasks on paper. Use the rubric to review your reasoning with a teacher or study partner; these tasks are not automatically scored.
Construct, test, and justify
On paper, draw a nonhorizontal segment CD. Construct its perpendicular bisector with a compass and straightedge, leaving the arcs visible. Choose a point R on your constructed line. Explain why RC = RD. Explain why a compass radius less than half CD cannot produce the two required intersections.
Review rubric and solution
- Two pairs of intersecting arcs use the same compass radius, and the straightedge line passes through their intersections.
- The argument identifies a midpoint and a right angle from congruence, rather than from measurement alone.
- The final equal-distance claim uses congruent right triangles (or the proved perpendicular-bisector theorem); the radius explanation uses the distance between circle centers.
- Repeat the construction in the example. Its SSS/SAS argument shows that the new line meets CD at midpoint M at right angles.
- For R different from M, right triangles RMC and RMD have equal legs RM and MC = MD and equal included right angles. SAS gives RC = RD. For R = M this is the midpoint property.
- If each radius r is less than CD/2, then 2r < CD. The circles cannot meet: a shared point would require CD ≤ r + r by the triangle inequality.
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