Make formal geometric constructions with compass and straightedge and other tools, and inscribe an equilateral triangle, a square and a regular hexagon in a circle

Learning goal: Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometry software): copying a segment, copying an angle, bisecting a segment, bisecting an angle, constructing perpendicular lines including the perpendicular bisector of a segment, and constructing a line parallel to a given line through a point not on the line; and construct an equilateral triangle, a square and a regular hexagon inscribed in a circle, justifying each construction with the theorems of this unit.

What makes a construction formal. A formal construction uses only operations whose results are exact by definition: drawing the line through two known points (straightedge), drawing the circle with a known center through a known point or with a known radius (compass), and marking the points where two drawn objects meet. Measuring with a ruler or protractor is not allowed, because a measurement is an approximation and carries no proof. The same constructions can be carried out with string (a taut string swings a circle), a reflective device (a transparent mirror shows where a reflection lands, which produces perpendiculars and bisectors), paper folding (a fold is a reflection, so folding A onto B creases the perpendicular bisector of AB) and dynamic geometry software (whose compass and line tools are exact). Whatever the tool, a construction is complete only when its steps are recorded and its result is justified by a theorem.

Copying a segment. Given segment AB and a ray from point C: set the compass to radius AB, draw the circle (or arc) about C with that radius, and mark D where it meets the ray. CD = AB by the definition of a circle, every point of which is at distance AB from C.

Bisecting a segment and constructing its perpendicular bisector. Given AB: draw circles about A and about B with the same radius r, chosen larger than half of AB so the circles meet; call the meeting points X and Y; draw line XY. Justification: XA = XB = r and YA = YB = r, so X and Y are both equidistant from A and B; by the perpendicular bisector theorem each lies on the perpendicular bisector of AB; two points determine a line, so XY is the perpendicular bisector. Its intersection M with AB is the midpoint. This single construction gives both "bisect a segment" and "perpendicular bisector". A fold that lands A on B produces the same line, because a reflection that swaps A and B is the reflection across the perpendicular bisector of AB.

Perpendicular through a point. For a point P on line ℓ: draw a circle about P meeting ℓ at two points A and B (so P is the midpoint of AB), then construct the perpendicular bisector of AB; it passes through P and is perpendicular to ℓ. For a point P not on ℓ: draw a circle about P large enough to meet ℓ at two points A and B, then construct the perpendicular bisector of AB. Since PA = PB, P lies on that bisector, and the bisector is perpendicular to ℓ. Both cases rest on the same theorem.

Bisecting an angle. Given angle with vertex V: draw a circle about V meeting the two sides at A and B, so VA = VB. Draw circles about A and B with a common radius greater than half of AB, so that they meet at a point C inside the angle. Draw ray VC. Justification: triangles VAC and VBC have VA = VB, AC = BC (equal radii) and VC = VC, so they are congruent by SSS, and the corresponding angles AVC and BVC are congruent: ray VC bisects the angle.

Copying an angle. Given an angle at V and a ray from point W: draw a circle about V meeting the sides at A and B, and the circle of the same radius about W meeting the ray at A′. Set the compass to the distance AB, draw the circle about A′ with that radius, and let it meet the circle about W at B′. Draw ray WB′. Justification: triangles VAB and WA′B′ have VA = WA′, VB = WB′ (equal radii) and AB = A′B′ (the copied distance), so they are congruent by SSS and the angle at W equals the angle at V.

Parallel through a point. Given line ℓ and point P not on it: draw any line through P meeting ℓ at Q; at P copy the angle that this transversal makes with ℓ at Q, placing the copy in the corresponding position at P; the new side through P is the required parallel. Justification: the transversal makes congruent corresponding angles with ℓ and with the new line, and congruent corresponding angles make the lines parallel (the converse of the corresponding angle theorem). An alternative uses two perpendiculars: construct the perpendicular from P to ℓ, then the perpendicular to that line at P; two lines perpendicular to the same line are parallel.

Regular hexagon inscribed in a circle. Given a circle with center O and radius r: mark any point A on the circle; with the compass still set to r, draw the arc about A meeting the circle at B; then the arc about B meeting the circle at C; continue to D, E, F. Joining the six points in order gives the hexagon. Justification: OA = OB = AB = r, so triangle OAB is equilateral and the angle AOB is 60°; six such angles fill exactly 360°, so the sixth arc returns to A, and the six chords are all r. With center (0, 0) and radius 2, the vertices are (2, 0), (1, √3), (−1, √3), (−2, 0), (−1, −√3), (1, −√3), and each side, for example from (2, 0) to (1, √3), has length √(1 + 3) = 2 = r.

Equilateral triangle and square inscribed in a circle. Joining every other vertex of the hexagon, A, C and E, gives an inscribed equilateral triangle: each side spans two 60° arcs, so the three sides are congruent chords, and with radius 2 each is √(9 + 3) = √12 ≈ 3.46, for example from (2, 0) to (−1, √3). For the square: draw any diameter through O, meeting the circle at A and C; construct the perpendicular bisector of AC (which passes through O because O is the midpoint); it meets the circle at B and D. ABCD is a square: the diagonals AC and BD are diameters, so they are congruent and bisect each other, which makes ABCD a rectangle (parallelogram with congruent diagonals), and they are perpendicular, so the four right triangles AOB, BOC, COD and DOA are congruent by SAS (legs equal to the radius, right angles at O), which makes the four sides congruent. With radius 2 the vertices are (2, 0), (0, 2), (−2, 0), (0, −2) and each side is √(4 + 4) = √8 ≈ 2.83.

The misconception. Students believe a construction that looks right is right, and skip the justification. A hexagon stepped off with a compass set slightly wider than the radius closes with a visible gap on the sixth step; the construction is exact only because the chord of a 60° arc equals the radius, and that is a theorem, not an observation. Record the steps, then write the reason each marked point is where you claim.

Worked, perpendicular bisector with coordinates: A(2, 1) and B(8, 5). The compass radius must exceed half of AB = √(36 + 16) = √52 ≈ 7.21, so half is about 3.61; choose radius 5. The circles (x − 2)² + (y − 1)² = 25 and (x − 8)² + (y − 5)² = 25 meet at two points X and Y that are equidistant from A and B. The perpendicular bisector is the line through the midpoint (5, 3) with slope −1 ÷ (4/6) = −3/2: y − 3 = (−3/2)(x − 5). The point (3, 6) is on it, since 6 − 3 = 3 and (−3/2)(3 − 5) = 3, and its distances to A and B are both √(1 + 25) = √26. Every point on the constructed line passes the same check.

Worked, bisecting an angle and justifying it: Angle with vertex V(0, 0), one side along the positive x-axis and the other along the ray through (0, 1), so the angle is 90°. Draw the circle of radius 4 about V: it meets the sides at A(4, 0) and B(0, 4). Draw circles of radius 4 about A and B; one meeting point inside the angle is C(4, 4), since CA = 4 and CB = 4. Ray VC is the line y = x, which makes 45° with each side. Justification: triangles VAC and VBC have VA = VB = 4, AC = BC = 4, VC common, so SSS gives ∠AVC ≅ ∠BVC, each 45°.

Worked, inscribing a regular hexagon and checking the sides: Circle with center O(0, 0) and radius 2. Start at A(2, 0). The arc of radius 2 about A meets the circle at B(1, √3): check OB = √(1 + 3) = 2 and AB = √((2 − 1)² + 3) = 2. Continue: C(−1, √3), D(−2, 0), E(−1, −√3), F(1, −√3). Each consecutive pair is 2 apart, for instance C to D: √(1 + 3) = 2, and F to A: √(1 + 3) = 2. The central angle AOB has cosine (2 × 1 + 0) ÷ (2 × 2) = 1/2, so it is 60°, and six of them make 360°.

Worked, inscribing a square and proving it is one: Same circle. Diameter through A(2, 0) and C(−2, 0). The perpendicular bisector of AC is the y-axis, which meets the circle at B(0, 2) and D(0, −2). ABCD has diagonals AC and BD of length 4 each, both bisected at O, so it is a parallelogram (diagonals bisect each other) with congruent diagonals, hence a rectangle; the diagonals are also perpendicular, so the sides are congruent: each is √(2² + 2²) = √8 ≈ 2.83. A rectangle with four congruent sides is a square.

Recording task: With compass and straightedge, carry out and record these constructions on one page: copy a segment of your choice; construct the perpendicular bisector of a segment AB of length 8 cm and mark its midpoint; bisect a 70° angle drawn with a protractor only for the starting angle; copy that 70° angle onto a new ray; construct the perpendicular to a line from a point not on it; and construct the parallel to a line through a point not on it by copying a corresponding angle. For each, write the numbered steps and a one-sentence justification naming the theorem (perpendicular bisector theorem, SSS, converse of the corresponding angle theorem). Then repeat the perpendicular bisector by paper folding and by software, and record whether all three methods place the midpoint at the same point.

Independent performance: Inscribe an equilateral triangle, a square and a regular hexagon in three circles of the same radius using compass and straightedge only, and then, in a coordinate model with center (0, 0) and radius 2, list the vertices of each figure and compute the side lengths (hexagon 2, square √8, triangle √12). Success criteria: every construction is done without measuring and the steps are numbered; each figure comes with a written justification that the sides are congruent (equilateral triangles about the center for the hexagon; diameters as congruent perpendicular diagonals for the square; equal arcs for the triangle); the coordinate side lengths match the claimed values; one construction is repeated with a second tool (string, reflective device, folding or software) with a note on what that tool makes easier or harder; and a closing paragraph explains why the sixth compass step of the hexagon lands exactly on the starting point.