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Lesson 1 of 12

Constructions and Tilings · Lesson 1 of 12

Eyes and the Perpendicular Bisector

“Discover how equal distances turn a symmetric eye design into an exact way to bisect a segment.”

Learning Objectives

• Construct symmetric arcs using equal distances. • Explain midpoint, bisection, and perpendicular bisector. • Justify the construction using SSS and SAS congruence. • Choose radii that produce usable intersections. • Investigate how changing the centres changes an eye design.

Eyes

An eye drawn freehand may look symmetric, yet its upper and lower curves can differ slightly. A compass lets us make the relationship exact. We begin with the two corners X and Y and look for centres from which both corners are equally far away.

A compass draws part of a circle, called an arc. Every point on that arc is the same distance from its centre. To draw an arc through both X and Y, its centre A must satisfy AX = AY. For a matching arc on the other side, choose B so that BX = BY. In the symmetric construction, all four distances are equal: AX = AY = BX = BY.

Draw intersecting arcs of equal radius from X and Y above the segment, and repeat below it. Call the intersections A and B. The points are at equal distances from X and Y because each lies on both equal-radius circles. Use A and B as centres to draw the two curves through the corners. Keep the supporting segment and construction arcs faint so the eye outline stands out.

XYABAX = AY = BX = BY
Equal distances behind the eye— A draws the lower arc and B draws the upper arc. Equal radii and symmetric centres make the two boundaries match.

Bisection and the Perpendicular Bisector

Join A to B and look at its intersection O with XY. Two things happen together: the segment is divided equally, and the crossing forms right angles. We need both facts to describe the new line correctly.

Definition
Bisection

Dividing a geometric quantity into two equal parts. Bisecting a segment produces two segments of equal length; bisecting an angle produces two equal angles.

Definition
Perpendicular bisector

A line that passes through the midpoint of a segment and meets the segment at 90°. The midpoint is the point that divides the segment into two equal lengths.

XYABOOXOY
The two properties of a perpendicular bisector— The triangles highlight the equal halves. A perpendicular alone need not pass through the midpoint.

Why the Construction Works

A neat drawing suggests the result, but it cannot establish that the result always holds. Congruence provides the reason. Congruent triangles have matching side lengths and angles, even when the drawing is turned or enlarged.

First compare triangles ABX and ABY. AX = AY and BX = BY come from the compass construction. AB is common to the two triangles. Their three corresponding sides are equal, so they are congruent by SSS, meaning side–side–side. Therefore ∠XAB = ∠YAB. Since O lies on AB between A and B, this also gives ∠XAO = ∠YAO.

Now compare triangles AOX and AOY. AX = AY, AO is common, and the included angles ∠XAO and ∠YAO are equal. These triangles are congruent by SAS, meaning side–angle–side. Corresponding parts therefore give OX = OY and ∠AOX = ∠AOY. The latter angles lie next to each other on a straight line, so their sum is 180°. Two equal angles with sum 180° are each 90°. Thus O is the midpoint and AB is perpendicular to XY.

Points Equidistant from the Endpoints

The construction reveals a useful location rule. Equidistant means equally far from two specified points. Once we know where all such points lie, we can find other centres and other ways to draw the same bisector.

Every point P on the perpendicular bisector has PX = PY. To see why, join P to both endpoints. The two right triangles have the same midpoint-to-endpoint distance, a common perpendicular side, and equal included right angles, so they are congruent by SAS.

Conversely, suppose PX = PY. Let M be the midpoint of XY and join P to M. Triangles PMX and PMY have equal corresponding sides by SSS: PX = PY, MX = MY, and common PM. The adjacent angles at M are equal and sum to 180°, so PM is perpendicular to XY. Thus P lies on its perpendicular bisector. If P is M itself, it already lies on that line. This explains why different pairs of equally distant arc centres all locate the same line.

Construction of Perpendicular Bisector

The argument gives a method using an unmarked straightedge and a compass. The straightedge draws a straight line; it does not need to measure the midpoint. The compass carries equal distances from one endpoint to the other.

  1. Draw the given segment XY.
  2. Choose a compass radius greater than half XY. From X and Y, draw arcs with that same radius so they intersect above XY at A.
  3. Repeat below XY to obtain B, keeping the two radii within this pair equal.
  4. Draw the line AB. Its intersection with XY is the midpoint, and AB is the perpendicular bisector.
Example — Bisect a 6 cm segment

Problem
Construct the perpendicular bisector of XY when XY = 6 cm.

  1. 1.Draw XY = 6 cm. Half its length is 3 cm, so choose a compass opening of 4 cm to obtain two intersections.
  2. 2.With centres X and Y and radius 4 cm, draw the crossing arcs above and below XY. Label the crossings A and B.
  3. 3.Join A and B. Equal distances from the endpoints place both on the perpendicular bisector.
  4. 4.The crossing O gives XO = OY = 3 cm. The lengths confirm the result, but the construction located O without measuring along XY.
Example — Different upper and lower radii

Problem
Can the upper pair use radius 5 cm and the lower pair radius 4 cm for a 6 cm segment?

  1. 1.Both radii exceed 3 cm, so both pairs can give intersections.
  2. 2.The upper point is 5 cm from each endpoint. The lower point is 4 cm from each endpoint. Each point is independently equidistant from X and Y.
  3. 3.Both lie on the same perpendicular bisector. Join them to draw that line. The radii of the upper and lower pairs need not match each other.

Exploring Different Radii

Changing a radius can change the appearance of the eye without changing the supporting bisector. This is a chance to separate a condition that is essential from one that was simply convenient in the first construction.

Example — Two intersections on the same side

Problem
How can two points above XY determine the perpendicular bisector?

  1. 1.Draw one pair of equal-radius arcs above XY and call the intersection A.
  2. 2.Use a different suitable radius for another pair above XY and obtain a distinct point C.
  3. 3.AX = AY and CX = CY, so both points lie on the perpendicular bisector. The line through A and C is the required line.
  4. 4.A single point is not enough to determine a unique line; the points must be distinct.
Common mistake

Within each standard pair, using different radii from X and Y destroys the guarantee of equal distances. Also, a radius smaller than half XY gives no intersections; exactly half gives only one touching point. Use a radius greater than half XY for two intersections.

For symmetric eye variations, choose centres on opposite sides of XY at equal distances from its midpoint. Their radii to X and Y then match. Moving the centres changes the curvature. A point equidistant from the endpoints alone does not guarantee symmetry between the two eye arcs unless the opposite centre is chosen to match. Use two perpendicular support directions to develop a four-petal design, keeping the corresponding arcs equal.

Centre offset: 50Centre offset: 100Centre offset: 160
Different curvature with the same corner separation— Each pair of centres is symmetric. Larger centre offsets make these connecting arcs flatter.
Four petals from symmetric arc pairs— Recreate the design using supporting lines first; then trace only the final curved boundary.

Check Your Understanding

Use these questions to check the conditions behind each method, as well as the result. For a diagram-based question, follow the labels and given information rather than judging by appearance.

Quiz

Quick check

Which pair of properties defines a perpendicular bisector?

Quick check

For a segment 8 cm long, which radius gives two intersections of equal-radius circles?

Quick check

Why are triangles ABX and ABY congruent in the standard construction?

Quick check

Which point must lie on the perpendicular bisector of XY?

Quick check

Must the radius used above XY equal the radius used below XY?

Practice Problems

Practice Problems
  1. Construct the perpendicular bisector of a 7 cm segment using a suitable radius. Explain your choice.
  2. A line crosses XY at 90° but divides it into 2 cm and 5 cm. Is it the perpendicular bisector?
  3. Explain why equal arcs of radius 3 cm cannot give two intersections for XY = 6 cm.
  4. Locate two distinct equidistant points on the same side of a segment and join them. Explain the result.
  5. Draw two symmetric eye shapes using the same endpoints but different matching centres.
  6. Recreate the four-petal design. Keep the supporting lines visible in a first version and trace the finished boundary in a second.

Choose a radius greater than 3.5 cm, for example 4.5 cm. Equal arcs from the endpoints locate two equidistant points; their line bisects the segment at 90°.

Key Takeaways

Key Takeaways

• A perpendicular bisector passes through a midpoint at 90°. • Equal-radius arcs locate points equidistant from the endpoints. • SSS followed by SAS explains the standard construction. • Equidistant points lie on the perpendicular bisector, and points on it are equidistant. • Use a radius greater than half the segment for two standard arc intersections. • Different pairs may use different radii while preserving equidistance.

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Next · Lesson 2

Construction of a 90° Angle at a Given Point