Lesson 1.1 · Unit 1 · G-CO.1

Starting from terms the subject refuses to define

Geometry is built on proof, and a proof has to start somewhere. If every term were defined using other terms, the definitions would either run forever or circle back on themselves. So three terms are left undefined on purpose, described but never defined, and everything else is built from them.

The method
  1. Point, line and plane are undefined terms. They are described, not defined: a point has position but no size, a line extends without end in two directions and has no thickness, a plane is a flat surface extending without end.
  2. Every other term in the course is defined from them, which is what makes a chain of proof possible rather than circular.
  3. A point is named by a capital letter; a line by two points on it with a double-headed arrow, \( \overleftrightarrow{AB} \), or by a single lowercase letter.
  4. A segment is the part between two endpoints, written \( \overline{AB} \), and a ray starts at one endpoint and continues through another, written \( \overrightarrow{AB} \).
  5. The order of letters matters for a ray and not for a line or segment. \( \overrightarrow{AB} \) starts at \( A \); \( \overrightarrow{BA} \) starts at \( B \), and they are different rays.
  6. Collinear points lie on one line; coplanar points lie in one plane. Any two points are collinear and any three are coplanar, so the terms only say something when applied to three or four.
  7. Two distinct lines intersect in at most one point, and two distinct planes that intersect do so in a line.
  8. Through any two points there is exactly one line, and through any three noncollinear points exactly one plane. These are postulates: accepted without proof because the subject has to begin somewhere.

Where students lose marks: using \( AB \) when \( \overline{AB} \) is meant. Without the bar, \( AB \) means the length, a number. \( \overline{AB} \cong \overline{CD} \) says two segments are congruent; \( AB = CD \) says two numbers are equal. Both are correct statements and a proof must use the right one.

Worked example

The problem. Points \( A \), \( B \) and \( C \) lie on one line with \( B \) between \( A \) and \( C \). Point \( D \) is not on that line. (a) Name every segment determined by these four points. (b) Is \( \overrightarrow{AB} \) the same as \( \overrightarrow{BA} \)? (c) How many planes contain all four points? (d) Explain why \( \overline{AB} \cong \overline{BC} \) and \( AB = BC \) say the same thing in different language.

Step one: count the segments in (a) systematically. A segment is determined by any two of the four points, so count pairs rather than trying to list them from a picture. Four points taken two at a time give six pairs.

Step two: list them. \( \overline{AB} \), \( \overline{AC} \), \( \overline{AD} \), \( \overline{BC} \), \( \overline{BD} \), \( \overline{CD} \). Six, as counted. Note that \( \overline{BA} \) is not a seventh: a segment has no direction, so \( \overline{AB} \) and \( \overline{BA} \) name the same segment.

Step three: answer (b) by checking the endpoint. A ray does have direction. \( \overrightarrow{AB} \) begins at \( A \) and passes through \( B \), continuing past it. \( \overrightarrow{BA} \) begins at \( B \) and passes through \( A \), continuing past it in the opposite direction. They are different rays.

Step four: make the difference concrete. Since \( B \) is between \( A \) and \( C \), the point \( C \) lies on \( \overrightarrow{AB} \), because going from \( A \) through \( B \) and onward reaches \( C \). But \( C \) does not lie on \( \overrightarrow{BA} \), which heads away from \( C \). One point settles the question.

Step five: set up (c) using the postulate. Three noncollinear points determine exactly one plane. Here \( A \), \( B \) and \( C \) are collinear, so they do not determine a plane by themselves; infinitely many planes contain a single line.

Step six: finish (c). Take any two of \( A \), \( B \), \( C \) together with \( D \). Since \( D \) is off the line, those three points are noncollinear and determine exactly one plane. That plane contains the line through \( A \), \( B \) and \( C \), since it contains two of its points, and it contains \( D \). So there is exactly one plane containing all four.

Step seven: answer (d) by identifying what each statement is about. \( \overline{AB} \cong \overline{BC} \) is a statement about two geometric objects: the segments are congruent, meaning one can be laid exactly on the other. \( AB = BC \) is a statement about two numbers: the lengths are equal.

Step eight: state the relationship and why the distinction is kept. The two statements are equivalent, since segments are congruent exactly when their lengths are equal, and a proof may move between them by the definition of congruent segments. They are kept separate because the reason column has to name which one a line is asserting. A step that adds lengths is using the arithmetic statement; a step that concludes two figures coincide is using the geometric one. Writing \( \overline{AB} = \overline{BC} \) mixes them and is marked wrong.

Practice · 10 problems

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

  1. Name the three undefined terms of geometry.
    Show the full solution

    Point, line and plane

  2. What does \( \overline{PQ} \) mean?
    Show the full solution

    The segment with endpoints \( P \) and \( Q \)

  3. What does \( PQ \) mean, without the bar?
    Show the full solution

    The length of that segment, a number

  4. Are \( \overrightarrow{XY} \) and \( \overrightarrow{YX} \) the same ray?
    Show the full solution

    They have different endpoints and point in opposite directions. No

  5. How many lines pass through two distinct points?
    Show the full solution

    Exactly one

  6. How many segments are determined by five points, no three collinear?
    Show the full solution

    A segment is determined by a pair of points, so count pairs. The first point can be chosen five ways and the second four, giving 20 ordered pairs, but a segment has no direction so each is counted twice: \( 20 \div 2 = 10 \). Checking by listing with points \( A \) through \( E \): \( AB, AC, AD, AE \) is four; \( BC, BD, BE \) is three; \( CD, CE \) is two; \( DE \) is one. Total \( 4 + 3 + 2 + 1 = 10 \). 10

  7. Three points lie on a line. How many planes contain all three?
    Show the full solution

    The postulate that three points determine exactly one plane requires them to be noncollinear, so it does not apply here. Picture the line as an axle: a plane containing it can be rotated about it to any position and still contains all three points. Infinitely many

  8. Explain why point, line and plane are left undefined rather than being defined carefully.
    Show the full solution

    Every definition explains one term using other terms. Those other terms then need definitions of their own, using still others. Only two endings are possible: the chain runs forever, or it comes back on itself and defines a term using something that was defined from it. Neither gives a foundation a proof can rest on. The resolution is to stop deliberately. A few terms are taken as understood without definition, described well enough to work with, and every other term in the subject is defined from them. That gives a finite chain with a bottom. The same structure appears in the statements: a few postulates are accepted without proof, and every theorem is proved from them. Undefined terms and postulates are the same idea applied to words and to claims. Defining every term would require an infinite or circular chain, so a few are taken as understood

  9. Point \( B \) is between \( A \) and \( C \). Which of \( \overrightarrow{AB} \), \( \overrightarrow{AC} \), \( \overrightarrow{BA} \), \( \overrightarrow{BC} \) name the same ray?
    Show the full solution

    A ray is determined by its endpoint and its direction, not by which second point is named. \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) both start at \( A \) and head toward \( B \) and \( C \), which lie the same way since \( B \) is between them. Same endpoint, same direction, so they are the same ray. \( \overrightarrow{BA} \) starts at \( B \) heading toward \( A \). \( \overrightarrow{BC} \) starts at \( B \) heading the opposite way toward \( C \). Same endpoint, opposite directions, so they are different rays. Together they make the whole line, and two such rays are called opposite rays. \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) are the same; \( \overrightarrow{BA} \) and \( \overrightarrow{BC} \) are opposite rays

  10. A student writes \( \overline{AB} = \overline{CD} \) in a proof. Explain what is wrong and give both correct versions.
    Show the full solution

    The equals sign relates numbers, and \( \overline{AB} \) is not a number; it is a segment, a set of points. Saying one set of points equals another would mean they are the same segment, which is almost never what is intended. Two correct statements are available, and they say different things in different language. \( \overline{AB} \cong \overline{CD} \) says the segments are congruent. The reason column would cite a theorem about figures. \( AB = CD \) says their lengths are equal. The reason column would cite an arithmetic property, and this is the form needed before adding or subtracting lengths. They are equivalent, and the definition of congruent segments is the theorem that licenses moving between them. A proof that needs to add lengths converts to the equation form first and says so. Use \( \overline{AB} \cong \overline{CD} \) for the figures or \( AB = CD \) for the lengths; an equals sign between two segments is not meaningful

1.2 Segments, distance and the midpoint →

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