Lesson 14: Geometric Constructions with Compass and Straightedge

Long before CAD software, builders laid out perfect right angles and exact divisions with two humble tools: a compass (draws circles and carries distances) and a straightedge (draws lines — *no measuring marks allowed*). A construction is a sequence of moves with these tools that is guaranteed correct by proof, not by measurement. This lesson covers the standard constructions and, just as importantly, *why* they work — every justification is a congruence argument from Lesson 3.

The basic moves and the core constructions

Everything is built from drawing lines through two points and circles with a chosen center and radius. The classic constructions:

• Copy a segment / copy an angle: the compass carries a distance without numbers.
• Perpendicular bisector of a segment: open the compass to more than half the segment's length; draw arcs from both endpoints above and below; the two intersection points define the perpendicular bisector.
• Angle bisector: arc from the vertex cutting both sides, then equal arcs from those two cut points meeting inside the angle; the ray from the vertex through that meeting point bisects the angle.
• Perpendicular from a point to a line, and a parallel line through a point (by copying a corresponding angle).
• Equilateral triangle on a given segment: circles of radius equal to the segment from each endpoint meet at the third vertex.
• Regular hexagon inscribed in a circle: step the radius around the circumference — six steps land exactly back at the start, because the side of an inscribed regular hexagon equals the radius.

The pattern to notice: constructions create pairs of congruent triangles, and the desired equality (of lengths, angles, or a 90° angle) falls out by CPCTC. A construction you cannot justify is just a drawing.

Worked example 1: Justifying the perpendicular bisector

Segment AB; from A and from B, arcs of the *same* radius r (more than half of AB) meet at P (above) and Q (below). Claim: line PQ is the perpendicular bisector of AB. Proof sketch: AP = BP = r and AQ = BQ = r (same compass opening), and PQ is shared, so △APQ ≅ △BPQ by SSS; hence ∠APM ≅ ∠BPM where M is where PQ crosses AB. Then in triangles APM and BPM: AP = BP, the angles at P are congruent, PM is shared, so △APM ≅ △BPM by SAS. CPCTC gives AM = BM — M is the midpoint — and ∠AMP = ∠BMP; those two angles form a linear pair summing to 180°, so each is 90°. PQ therefore bisects AB at right angles. ∎ Every step used only the construction's own guarantees.

Worked example 2: The perpendicular bisector as an equation (coordinate check)

Take A(0, 0) and B(6, 2). The construction should produce the line through the midpoint (3, 1) with slope the negative reciprocal of AB's slope. Slope of AB is (2 − 0)/(6 − 0) = 1/3, so the bisector has slope −3 and equation y − 1 = −3(x − 3). Check a second point on it: x = 2 gives y = 4; distance from (2, 4) to A is √(4 + 16) = √20, and to B is √(16 + 4) = √20. Equal — every point of the perpendicular bisector is equidistant from A and B, which is exactly the property the construction exploits in reverse.

Worked example 3: Equilateral triangle and the 60° angle

On segment AB of length 5, draw the circle centered at A with radius 5 and the circle centered at B with radius 5. They intersect at C with AC = 5 (radius of A's circle) and BC = 5 (radius of B's circle), so AB = BC = CA = 5: an equilateral triangle, by SSS applied to the radii. Each angle is therefore 60°, and this construction is the source of exact 60° and (by bisection) 30° angles in classical geometry. Bisect one of those angles with the angle-bisector construction and you own 30°; bisect again for 15°. A small kit of exact angles, built from nothing but circles.

Worked example 4: Bisecting an angle, justified

To bisect angle with vertex O: an arc centered at O cuts the sides at X and Y, so OX = OY (radii). Equal arcs from X and Y meet at Z inside the angle, so XZ = YZ (same compass opening) and OZ is shared. Then △OXZ ≅ △OYZ by SSS, so ∠XOZ ≅ ∠YOZ by CPCTC — the ray OZ genuinely splits the angle in two equal parts. The same SSS skeleton justified the perpendicular bisector; recognising the repeating skeleton is the real lesson.

Geometry at work

Construction geometry is not a museum piece. Surveyors lay out perpendicular offsets from a baseline with the modern descendants of these moves; carpenters still strike arcs with a compass (or a string and two nails — a garden-scale compass) to bisect boards and lay out arches. CNC programmers and CAD designers use "construction geometry" — circles and lines that exist only to locate features exactly. And the equidistance property of the perpendicular bisector decides real disputes: the set of points closer to one town than another is bounded by the perpendicular bisector of the segment joining them, which is how mobile coverage and service territories are modelled.

Common mistakes

• Measuring with the straightedge. Constructions forbid it; the guarantee comes from congruent triangles, not from a ruler.
• Changing the compass opening mid-move. The perpendicular bisector fails if the arcs from A and B use different radii; label openings and keep them fixed where the construction demands.
• Choosing an arc radius too small. For the perpendicular bisector the radius must exceed half the segment or the arcs never meet.
• Asserting the result without the justification. "It looks perpendicular" is not a construction proof; name the congruent triangles.

Exam Experience

1. In the standard construction, two arcs of equal radius are drawn from both endpoints of a segment and intersect above and below. What has been constructed?

• A) An angle bisector
• B) The perpendicular bisector of the segment
• C) A parallel line
• D) An equilateral triangle

Answer: B.

2. What is the side length of a regular hexagon inscribed in a circle of radius 9?

• A) 4.5
• B) 9
• C) 18
• D) 9√3

Answer: B. The side of an inscribed regular hexagon equals the radius.

3. The justification for the angle-bisector construction rests mainly on which congruence pattern?

• A) ASA, then CPCTC
• B) SSS, then CPCTC
• C) HL, then CPCTC
• D) AAA, then CPCTC

Answer: B. Radii and the shared ray give three pairs of congruent sides; CPCTC delivers the congruent angles. (AAA never proves congruence.)

Constructed response. Give the steps of the construction that bisects a given segment AB and explain, with a proof sketch, why the constructed line is perpendicular to AB at its midpoint. Your explanation must reference at least one pair of congruent triangles and the linear pair formed at the crossing point.

Scoring guide (4 points):

• 1 point: Steps: compass opened to more than half of AB; arcs from A and from B with the same opening intersect at P and Q; line PQ is drawn and crosses AB at M.
• 1 point: △APQ ≅ △BPQ by SSS (AP = BP, AQ = BQ as equal radii; PQ common), or an equivalent first congruence.
• 1 point: From ∠APM ≅ ∠BPM (CPCTC) and SAS, △APM ≅ △BPM, so AM = BM by CPCTC — the line bisects AB.
• 1 point: ∠AMP and ∠BMP are congruent (CPCTC) and form a linear pair summing to 180°, so each is 90° — the line is perpendicular to AB at its midpoint.

Going deeper: Inscribing a square, and the limits of construction

To inscribe a square in a given circle: draw any diameter, then construct the perpendicular bisector of that diameter (Lesson 14's core move) to get a second diameter at right angles; join the four endpoints. Why is the result a square? The diagonals of the quadrilateral are the two diameters: they are congruent (both are diameters), they bisect each other (both pass through the center), so the quadrilateral is a rectangle (a parallelogram with congruent diagonals); and the diagonals are perpendicular, which makes a rectangle a square. Three theorems from earlier lessons assemble the justification.

It is equally important to know what constructions *cannot* do. Greek geometers posed three famous problems — trisecting an arbitrary angle, doubling a cube, and squaring a circle — and all three were proved impossible with compass and straightedge in the 19th century (the proofs need algebra beyond this course). Angle bisection is possible; trisection, in general, is not. Knowing the boundary of a method is part of mastering it.

Locus thinking: a construction is a set of points

Each construction secretly describes a locus — the set of all points satisfying a condition. The perpendicular bisector of AB is the locus of points equidistant from A and B (Lesson 14's coordinate check verified this for one point; the construction guarantees it for all). An angle bisector is the locus of points equidistant from the angle's two sides. A circle is the locus of points at distance r from a center. This viewpoint solves "find the point that..." problems elegantly: a point equidistant from two towns *and* lying on a given road is where the perpendicular bisector meets the road — two loci, one intersection. Fair-division boundaries (the set of points closer to school X than school Y) are perpendicular bisectors, which is why Lesson 5's midpoint computation keeps reappearing in logistics and politics alike. When a construction problem feels mysterious, ask "what condition does each drawn line or arc enforce?" — the arcs enforce equal distances, the lines enforce the loci, and the intersection point satisfies everything at once.