state precise definitions of segment, angle, circle, perpendicular and parallel lines, and construct an equilateral triangle and a perpendicular bisector while explaining why each construction works.
Instruction
Every subject has to start somewhere. Geometry starts with a few ideas it does not define: point, line, distance along a line and distance around a circular arc. Everything else is defined in terms of these, and those definitions are what proofs are allowed to use.
Precise definitions.
- A line segment AB is the points A and B and all points of line AB between them.
- A circle with center O and radius r is the set of all points at distance r from O.
- An angle is two rays with a common endpoint, the vertex.
- Two lines are perpendicular if they meet to form right angles (90°).
- Two lines in a plane are parallel if they never meet.
Notice how much work the word all does in the circle definition. It is not enough that the radius points are at distance r; every point at distance r must be on the circle. That is exactly what makes constructions work.
Construction 1: an equilateral triangle (Euclid, Elements I.1). Given segment AB: draw the circle with center A through B, and the circle with center B through A. Call one intersection point C. Draw AC and BC.
Why it works. C is on the circle centered at A, so AC = AB (definition of circle). C is also on the circle centered at B, so BC = BA. Therefore AC = AB = BC, and the triangle is equilateral. Notice that the argument never says the triangle looks equilateral.
Construction 2: the perpendicular bisector. Given AB: draw the circle centered at A through B and the circle centered at B through A. They meet at two points, P and Q. The line PQ is the perpendicular bisector of AB.
Why it works. By the same reasoning, PA = PB and QA = QB, so P and Q are each equidistant from A and B. In Unit 2 you will prove that the points equidistant from A and B are exactly the points of the perpendicular bisector. For now, notice that APBQ has four equal sides: it is a rhombus, and its diagonals cross at right angles and bisect each other.
Circles inside circles. The same tool builds regular polygons. If you mark off the radius six times around a circle, you land back where you started, because each chord equal to the radius makes an equilateral triangle with the center, and 6 × 60° = 360°. Joining the six points gives a regular hexagon; joining every other point gives an equilateral triangle. Two perpendicular diameters give the four vertices of an inscribed square.
Tools. Do each construction twice: once with compass and straightedge, once in GeoGebra using only the Circle with Center through Point, Line and Intersect tools. Drag the original points in GeoGebra. If the construction is correct, it survives the drag; if you eyeballed a point, it breaks. That is the software version of a proof.
Formative check
Work through these before moving on. They are not graded — they tell you, and your teacher, whether the standard below has landed yet.
You set a compass to the radius of a circle and step it around the circle, marking points. How many steps bring you back to the start?
Which definition of a circle is precise enough to use in a proof?
In the equilateral-triangle construction, AC = BC because the triangle looks symmetric.
Construct a regular hexagon inscribed in a circle in GeoGebra (or with compass and straightedge). List each step, then write a justification that each side of the hexagon equals the radius, using only definitions.
Simulation & tools
Construct and drag-test an equilateral triangle
Open a blank GeoGebra Geometry app. Use only these tools: Segment, Circle with Center through Point, Intersect and Polygon.
- Draw segment AB.
- Draw the circle with center A through B, then the circle with center B through A.
- Use Intersect to mark one crossing point C, then use Polygon to draw triangle ABC.
- Use the Distance or Length tool to display AB, BC and CA.
- Drag A and B around the screen. Record what happens to the three lengths.
- Now draw a second triangle by placing a free point D where it looks right above AB. Drag A again and record what happens to that triangle.
Answer: Why does the constructed triangle stay equilateral while the eyeballed one does not? Name the definition that guarantees AC = AB.
Without a device: do the same construction with compass and straightedge on paper, then redraw it starting from a segment of a different length and measure all three sides with a ruler.
Open GeoGebra ↗ · GeoGebra · free for non-commercial use
How confident are you that you can state precise definitions and justify an equilateral-triangle and a perpendicular-bisector construction?
Practice
Work these on paper or in your notebook, then open Check your answer. Aim for all of Fluency and Application; try at least one Challenge.
Printable version: this unit’s practice workbook (PDF)
Fluency
Build speed and accuracy with the core skill.
- Name the four undefined notions that geometry in this course starts from.
Check your answer
Answer: Point, line, distance along a line, and distance around a circular arc. - Complete the definition precisely: a circle with center O and radius r is …
Check your answer
Answer: the set of all points in the plane at distance r from O. - Define perpendicular lines and parallel lines precisely.
Check your answer
Answer: Perpendicular lines meet to form right (90°) angles. Parallel lines are lines in the same plane that never meet. - Why can “a round shape” not be used as a definition of a circle in a proof?
Check your answer
Answer: It names no center or radius, so it cannot justify a statement such as AC = AB.A usable definition must let you conclude facts about lengths; only the center-and-radius definition does. - In the equilateral-triangle construction on a segment AB of length 5 cm, find AC and BC and give the reason for each.
Check your answer
Answer: AC = 5 cm and BC = 5 cm.C is on the circle with center A through B, so AC = AB; it is on the circle with center B through A, so BC = BA. - A regular hexagon is inscribed in a circle of radius 4 cm by stepping the radius around the circle. Find its perimeter.
Check your answer
Answer: 24 cmEach side equals the radius, 4 cm, and there are six sides. - Joining every other vertex of that hexagon gives an equilateral triangle inscribed in the same circle (radius 4 cm). Find its side length exactly and to the nearest hundredth.
Check your answer
Answer: 4√3 ≈ 6.93 cmEach side is the long diagonal of a pair of adjacent equilateral triangles of side 4: twice the altitude 2√3. - A square is inscribed in a circle of radius 5 cm using two perpendicular diameters. Find the side and area of the square.
Check your answer
Answer: side 5√2 ≈ 7.07 cm; area 50 cm2Each side is the hypotenuse of a right isosceles triangle with legs 5.
Application
Use the skill in context. Show your reasoning.
- (Illustrative map.) Two schools sit at A(1, 2) and B(7, 6) on a grid in kilometers. A bus hub must be equidistant from both schools and lie on the highway y = 0. Where should it go?
Check your answer
Answer: (20/3, 0) ≈ (6.67, 0)Points equidistant from A and B lie on the perpendicular bisector of AB. Midpoint (4, 4); slope of AB = 4/6 = 2/3, so the bisector has slope −3/2: y − 4 = −3/2(x − 4). Setting y = 0 gives x = 20/3. Check: both squared distances are 325/9. - A hexagonal community garden is inscribed in a circle of radius 6 m. How much fencing does the hexagon need, and how much more would a circular fence need?
Check your answer
Answer: 36 m for the hexagon; about 1.70 m more for the circleHexagon: 6 × 6 = 36 m. Circle: 2π(6) = 12π ≈ 37.70 m. - A designer draws a 12 cm segment, constructs an equilateral triangle on it, and then constructs the perpendicular bisector of the base. How tall is the triangle?
Check your answer
Answer: 6√3 ≈ 10.39 cmThe bisector passes through the apex and splits the base into 6 and 6; the height is √(122 − 62) = √108 = 6√3.
Challenge
Stretch problems. Expect to think before you write.
- Explain why the perpendicular-bisector construction of Lesson 1.1 fails if both circles are drawn with a radius less than half of AB.
Check your answer
Answer: The circles do not meet, so there are no intersection points P and Q to join.Any point on both circles would be within r of A and of B, so AB ≤ 2r. If 2r < AB, no such point exists. - Using only compass and straightedge, describe how to construct a 30° angle, and justify each step.
Check your answer
Answer: Construct an equilateral triangle (each angle 60°), then bisect one of its angles.Equilateral triangle: angles equal by symmetry and sum to 180°, so each is 60°. The angle-bisector construction splits it into two 30° angles.
Review
Keep earlier skills sharp.
- Solve: −3x + 10 = 37
Check your answer
Answer: x = −9Subtract 10 from both sides: −3x = 27. Divide by −3. - Solve: −6x + 5 = 11
Check your answer
Answer: x = −1Subtract 5 from both sides: −6x = 6. Divide by −6. - Find the slope of the line through (0, −4) and (0, 6).
Check your answer
Answer: Undefined (vertical line)The run is 0; division by zero is undefined. - A right triangle has legs 8 and 3. Find the hypotenuse exactly.
Check your answer
Answer: √73c2 = 82 + 32 = 73.