Lesson 1.6 · Unit 1 · G-CO.12, G-CO.13

Drawings that are guaranteed rather than approximate

A construction uses only a compass and an unmarked straightedge, and the restriction is the point. Anything a construction produces is exactly right, not close, and it can be proved so. Each construction in this lesson is followed by the argument for why it works, because a construction without that argument is just a picture.

The method
  1. A construction uses a compass and an unmarked straightedge only. No ruler measurements, no protractor, no eyeballing.
  2. The compass does two things: it draws a circle of a chosen radius, and it transfers a length from one place to another without measuring it.
  3. The straightedge does one thing: it draws the line through two points already constructed.
  4. Every construction works because the compass guarantees equal radii, which is the fact every proof of a construction ultimately uses.
  5. Copying a segment: draw a ray, set the compass to the original length, and mark that distance from the endpoint.
  6. Bisecting a segment: from each endpoint draw arcs of the same radius, more than half the segment's length, on both sides; the line through the two crossing points is the perpendicular bisector.
  7. Bisecting an angle: arc from the vertex to cut both sides, then equal arcs from those two points; the ray from the vertex through the crossing point is the bisector.
  8. Describe each step by saying where the compass point goes and what width it is set to, since that is what makes the instruction followable.

Where students lose marks: changing the compass width between steps that require it to stay fixed. In the segment bisection, both arcs must use the same radius; if they differ, the crossing points are not equidistant from both endpoints and the line through them is not the perpendicular bisector.

Worked example

The problem. Construct the perpendicular bisector of a given segment \( \overline{AB} \), then prove that the construction works.

Step one: set the compass. Open the compass to a width greater than half of \( AB \). The exact width does not matter, but it must exceed half the length, or the arcs will not reach each other and will not cross.

Step two: draw the first pair of arcs. Place the compass point on \( A \) and draw an arc above the segment and another below it, keeping the same width.

Step three: draw the second pair without changing the width. Move the compass point to \( B \), leaving the opening exactly as it was, and draw arcs above and below that cross the first pair. Call the crossing points \( P \) above and \( Q \) below.

Step four: draw the line. Lay the straightedge through \( P \) and \( Q \) and draw \( \overleftrightarrow{PQ} \). The claim is that this is the perpendicular bisector of \( \overline{AB} \).

Step five: begin the proof by recording what the compass guarantees. The compass width was never changed, so all four arcs have the same radius. Since \( P \) is on an arc centered at \( A \) and also on an arc centered at \( B \), both of that radius, \( PA = PB \). The same argument gives \( QA = QB \).

Step six: identify the triangles. Draw \( \overline{PA} \), \( \overline{PB} \), \( \overline{QA} \) and \( \overline{QB} \). Then \( \triangle PAQ \) and \( \triangle PBQ \) share the side \( \overline{PQ} \), and by step five \( PA = PB \) and \( QA = QB \).

Step seven: conclude congruence and then the bisection. The two triangles have three pairs of congruent sides, so they are congruent by SSS, which lesson 5.2 proves. Therefore \( \angle APQ \cong \angle BPQ \) by corresponding parts. Now \( \triangle APM \) and \( \triangle BPM \), where \( M \) is where \( \overline{PQ} \) meets \( \overline{AB} \), have \( PA = PB \), the congruent angles just found, and the shared side \( \overline{PM} \). They are congruent by SAS, so \( AM = BM \) and \( M \) is the midpoint.

Step eight: finish with perpendicularity. From the same congruence, \( \angle AMP \cong \angle BMP \). These two angles form a linear pair, so they are supplementary. Two angles that are both congruent and supplementary are right angles, by the small theorem proved at the end of lesson 1.4. So \( \overleftrightarrow{PQ} \perp \overline{AB} \). Both conditions of the definition now hold, so the constructed line is the perpendicular bisector. Every step of the argument traced back to one fact: the compass width never changed, so certain distances are equal. That is why the restriction to compass and straightedge is not an arbitrary game.

Practice · 10 problems

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

  1. What two tools are allowed in a construction?
    Show the full solution

    A compass and an unmarked straightedge

  2. Why is the straightedge unmarked?
    Show the full solution

    So it cannot be used to measure, only to draw a line through two points

  3. In bisecting a segment, must the compass width exceed half the length?
    Show the full solution

    Otherwise the arcs never meet. Yes

  4. What does a compass guarantee that a ruler does not?
    Show the full solution

    That two distances are exactly equal, without measuring either

  5. Describe the first step in bisecting an angle.
    Show the full solution

    Place the compass point on the vertex and draw an arc crossing both sides

  6. Describe the full construction for copying an angle.
    Show the full solution

    Given \( \angle A \) and a ray with endpoint \( D \). Place the compass point on \( A \) and draw an arc crossing both sides of the angle at points \( B \) and \( C \). Without changing the width, place the compass point on \( D \) and draw a large arc crossing the new ray at a point \( E \). Set the compass to the width \( BC \), the distance between the two crossing points of the first arc. Place the compass point on \( E \) and draw an arc crossing the large arc at a point \( F \). Draw \( \overrightarrow{DF} \). Then \( \angle FDE \cong \angle A \). It works because \( \triangle ABC \) and \( \triangle DEF \) have three pairs of congruent sides by construction, so they are congruent by SSS, and corresponding angles of congruent triangles are congruent. Arc across both sides, copy that arc at the new vertex, transfer the chord width, draw the ray

  7. Describe how to construct a perpendicular to a line through a point on that line.
    Show the full solution

    Let the point be \( P \) on line \( \ell \). Place the compass point on \( P \) and draw arcs crossing \( \ell \) on both sides, at points \( A \) and \( B \). By construction \( PA = PB \), so \( P \) is now the midpoint of \( \overline{AB} \). Now construct the perpendicular bisector of \( \overline{AB} \): widen the compass, draw arcs from \( A \) and from \( B \) that cross above the line, and draw the line from that crossing point through \( P \). That line is perpendicular to \( \ell \) and passes through \( P \), as required. The construction reduces to the one already proved, which is the usual pattern: later constructions are built from earlier ones rather than from scratch. Mark equal distances either side of \( P \), then bisect that segment

  8. Explain why a construction is closer to a proof than to a drawing.
    Show the full solution

    A drawing is as accurate as the hand that made it and the instrument that measured it. Measuring an angle with a protractor and drawing its bisector produces something approximately right, limited by the protractor's precision and by eyesight. A construction makes no measurements at all. It produces equalities by the compass, which guarantees that two distances are the same because the opening never changed, and then it derives the result from those equalities. The output is exactly correct as a matter of logic, not approximately correct as a matter of care. That is why every construction in this course is followed by its proof. The construction is the claim and the proof is the justification, exactly as elsewhere in geometry. A student who can perform the steps but cannot say why they work has learned the drawing and not the mathematics. It produces an exact result guaranteed by an argument, rather than an approximate one limited by measurement

  9. A student bisects a segment but changes the compass width between the arcs from \( A \) and the arcs from \( B \). What goes wrong?
    Show the full solution

    The whole proof rests on the crossing points being equidistant from \( A \) and from \( B \). If the width changes, the crossing point \( P \) satisfies \( PA = r_1 \) and \( PB = r_2 \) with \( r_1 \neq r_2 \), so \( PA \neq PB \). Without that equality, \( \triangle PAQ \) and \( \triangle PBQ \) are not congruent, nothing follows about the angles, and the line through the crossing points is neither perpendicular to \( \overline{AB} \) nor through its midpoint. Geometrically the line will still cross the segment, but at the wrong place and the wrong angle, and it will look close enough to pass a glance. That is what makes the error dangerous: the failure is invisible in the drawing and only the argument reveals it. The crossing points are no longer equidistant from both endpoints, so the line is neither perpendicular nor a bisector

  10. Describe how to construct an equilateral triangle on a given segment, and prove it is equilateral.
    Show the full solution

    Given \( \overline{AB} \). Set the compass to the width \( AB \). Place the point on \( A \) and draw an arc above the segment. Without changing the width, place the point on \( B \) and draw a second arc crossing the first at a point \( C \). Draw \( \overline{AC} \) and \( \overline{BC} \). Proof. The compass was set to \( AB \) and never changed. \( C \) lies on the arc centered at \( A \) with radius \( AB \), so \( AC = AB \). \( C \) also lies on the arc centered at \( B \) with the same radius, so \( BC = AB \). By the transitive property, \( AC = BC \), and all three sides have length \( AB \). A triangle with three congruent sides is equilateral by definition, so \( \triangle ABC \) is equilateral. This is the first proposition in Euclid's Elements, and it is a good example of how little machinery a construction proof needs: one fixed compass width and the transitive property. Two arcs of radius \( AB \) centered at \( A \) and \( B \); all three sides equal \( AB \) because the compass never changed

← 1.5 Perpendicularity, bisectors and proof-ready definitions1.7 Polygons and the vocabulary of the year →

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