Chapter1
    Euclid’s Elements, Book I
    (constructions)
                              102                                      Euclid’s Elements, Book I (constructions)
                              1.1      Theuseofrulerandcompass
                              Euclid’s Elements can be read as a book on how to construct certain geometric
                              figures efficiently and accurately using ruler and compass, and ascertaining the
                              validity. The first three postulates before Book I are on the basic use of the ruler
                              and the compass.
                              Postulate 1. To draw a straight line from any point to any point.
                                  With a ruler (straightedge) one connects two given points A and B to form the
                              line (segment) AB, and there is only one such line. This uniqueness is assumed, for
                              example, in the proof of I.4.
                              Postulate 2. To produce a finite straight line continuously in a straight line.
                                  Given two points A and B, with the use of a ruler one can construct a point C
                              so that the line (segment) AC contains the point B.
                                  Thefirsttwopostulatescanbecombinedintoasingleone: throughtwodistinct
                              points there is a unique straight line.
                              Postulate 3. To describe a circle with any center and distance.
                                  This distance is given by a finite line (segment) from the center A to another
                              point B. With the use of a collapsible compass, one constructs a circle with given
                              center A to pass through B. We denote this circle by C(A,B). Euclid I.2 shows
                              how to construct a circle with a given center and radius equal to a given line (seg-
                              ment).
                              Definition (I.20). Of trilateral figures,
                              an equilateral triangle is that which has its three sides equal,
                              an isosceles triangle that which has two of its sides alone equal, and
                              a scalene triangle that which has its three sides unequal. 1
                              Euclid (I.1). On a given finite straight line to construct an equilateral triangle.
                                                                         C
                                                             D      A                 E
                                                                              B
                              Given: Points A and B.
                              To construct: Equilateral triangle ABC.
                              Construction: Construct the circles C(A,B) and C(B,A) to intersect at a point C.
                              ABCisanequilateral triangle.
                                 1Euclid seems to take isosceles and scalene in the exclusive sense. But it is more convenient
                              to take these in the inclusive sense. An isosceles triangle is one with two equal sides, so that an
                              equilateral triangle is also isosceles.
                    1.1 The use of ruler and compass                                                   103
                        Thesecondproposition is on the use of a collapsible compass to transfer a seg-
                    ment to a given endpoint.
                    Euclid (I.2). To place a straight line equal to a given straight line with one end at
                    agiven point.
                                                           A    D
                                                                      C
                                                                B
                                                    L
                                                               G
                    Given: Point A and line BC.
                    To construct: Line AL equal to BC.
                    Construction: (1) An equilateral triangle ABD,                                    [I.1]
                    (2) the circle C(B,C).
                    (3) Extend DB to intersect the circle at G.
                    (4) Construct the circle C(D,G) and (5) extend DA to intersect this circle at L.
                    AL=DLŠDA=DGŠDB=BG=BC.
                        Therefore the circle C(A,BC) can be constructed using a collapsible compass.
                    Euclid (I.3). Given two unequal straight lines, to cut off from the greater a straight
                    line equal to the less.
                    Euclid (I.9). To bisect a given rectilineal angle.
                                                             A
                                                        D           E
                                                              F
                                                      BC
                    Given: Angle BAC.
                    To construct: Line AF bisecting angle ABC.
                    Construction: (1) Choose an arbitrary point D on AB.
                    (2) Construct E on AC such that AD = AE.                                          [I.3]
                    (3) Construct an equilateral triangle DEF on DE (so that F and A are on opposite
                    sides of DE).                                                                     [I.1]
                    Theline AF is the bisector of the angle BAC.
                              104                                      Euclid’s Elements, Book I (constructions)
                              Euclid (I.10). To bisect a given finite straight line.
                                                                         C
                                                                A        D         B
                              Given: Line segment AB.
                              To construct: The midpoint of AB.
                              Construction: (1) An equilateral triangle ABC,                                      [I.1]
                              (2) the bisector of angle ACB                                                       [I.9]
                              (3) to meet AB at D.
                              AD=DB.
                              1.2      Perpendicular lines
                              Definition (I.10). When a straight line set up on a straight line makes the adjacent
                              angles equal to one another, each of the equal angles is right,
                              andthestraightlinestandingontheotheriscalledaperpendiculartothatonwhich
                              it stands.
                              Postulate 4. That all right angles are equal to each other.
                              Euclid (I.11). To draw a straight line at right angles to a given straight line from a
                              given point on it.
                                                                        F
                                                           AB
                                                                  D     C     E
                              Given: A straight line AB and a point C on it.
                              To construct: A line CF perpendicular to AB.
                              Construction: (1) Take an arbitrary point D on AC, and construct E on CB such
                              that DC = CE.
                              (2) Construct an equilateral triangle DEF.                                          [I.1]
                              Theline CF is perpendicular to AB.
                              Proof: In triangles DCF and ECF,
                                DC=EC,                byconstruction
                                CF=CF,
                                DF =EF                sides of equilateral triangle
                                △DCF≡ECF SSS
                                ∠DCF =∠ECF
                                CF⊥AB.