Let’s start with some very basic constructions.
Definition 12.1.1. A perpendicular bisector of a line segment is a line that is perpendicular to the line segment and goes through the middle of the line segment.
Theorem 12.1.2. Given any line segment, its perpendicular bisector can be con- structed.
Proof. Given a line segment AB, as shown in Figure12.1, put the point of the compass at Aand open the compass to radius the length ofAB. Letr equal the length ofAB. Then draw the circle with center atAand radiusr. Similarly, draw the circle with center atBand radiusr. The two circles will intersect at points,C andD, as indicated in Figure12.1. Take the straightedge and draw the line segment fromC toD. We claim thatCDis a perpendicular bisector ofAB.
A E B
C
D
Fig. 12.1 Constructing the perpendicular bisector of a line segment
To prove this, label the point of intersection ofCDandAB asEand then draw the line segmentsAC,CB,BD, andDA. We must prove thatAE DEBand that
†CEA(and/or any of the other three angles atE) is a right angle. First note that AC,CB,BD, andDAall have the same length,r, since they are all radii of the two circles of radiusr. Thus, triangleACDis congruent to triangleDCB, since the third side of each isCDand they therefore agree by side-side-side (11.1.8). It follows that†ACE D †BCE. Thus, triangleACEis congruent to triangleBCEby side- angle-side (11.1.2). Therefore,AEDEB. Moreover,†AEC D †BEC, so, since those two angles sum to a straight angle, each of them is a right angle. ut Definition 12.1.3. An angle bisector is a line from the vertex of the angle that divides it into two equal subangles.
Theorem 12.1.4. Given any angle, its bisector can be constructed.
12.1 Constructions with Straightedge and Compass 129
A
B
C
E F
G
Fig. 12.2 Constructing the bisector of an angle
Proof. Consider an angle ABC, as pictured in Figure 12.2, and draw a circle centered at B that intersects both BA and BC. Label the points of intersection of the circle withAB and withBC asEandF, respectively. Letrbe the distance fromEtoF. Use the compass to draw a circle of radiusrcentered atEand a circle of radiusr centered atF. These two circles intersect in some pointG within the angleABC, as shown in Figure12.2. Use the straightedge to draw the line segment connectingBtoG. We claim that this line segment bisects the angleABC.
Draw the linesEGandF G. We prove that triangleBEGis congruent to triangle BF G. Note thatBEDBF, since they are both radii of the original circle centered atB. Note also thatEG D F G, since they are each radii of circles with radiusr. Since triangleBEGand triangleBF G share sideBG, it follows from side-side- side (11.1.8) that the two triangles are congruent. Thus,†EBGD †FBGandBG
is a bisector of angleABC. ut
Theorem 12.1.5. Any given line segment can be copied using only a straightedge and compass.
A B C D
Fig. 12.3 Copying a line segment
Proof. Suppose a line segmentAB is given, as pictured in Figure 12.3, and it is desired to copy it on another line. Choose any pointC on the other line, and then open the compass to a radius the length ofAB. Put the point of the compass atC and draw any portion of the resulting circle that intersects the other line. Label the
point of intersectionD. ThenCDis copy ofAB. ut
Theorem 12.1.6. Any given angle can be copied using only a straightedge and compass.
130 12 Constructability
A
B C G
D
E H
I
Fig. 12.4 Copying an angle
Proof. Let an angleABC be given, as in Figure12.4. We construct an angle equal to†ABC with vertexG on any other line. To do this, draw any arc of any circle (of radius, say,r) centered atBthat intersects bothBAandBC. Label the points of intersectionDandE. Draw the circle of radiusrcentered atG. UseH to label the point where that circle intersects the line containingG. Then adjust the compass to be able to make circles of radiusDE. Put the point of the compass at H and draw a portion of the circle that intersects the circle centered atG; call that point of intersectionI. Draw line segments connectingDtoEandItoH.
ThenIH DDE, sinceIH is a radius of a circle with radiusDE. Also draw the line segmentGI. The lengths ofBD,BE,GI, andGHare all equal tor. It follows by side-side-side (11.1.8) that triangleBDE is congruent to triangle GIH. Thus,
†I GHis a copy of†ABC. ut Corollary 12.1.7. If the angles˛andˇare constructed, then:
(i) the angle˛Cˇcan be constructed, and
(ii) for every natural numbern, the anglen˛can be constructed.
A
B α C
D
E β F
D
E β F
α
Fig. 12.5 Constructing the sum of two angles
Proof. (i) Let the angles˛andˇbe given, as pictured in Figure12.5. To construct the angle˛Cˇ, simply copy the angle˛ with one side DE and the other side outside the original angleˇ, as shown in the third diagram in Figure12.5.
(ii) This clearly follows from repeated application of part (i), starting with angles
˛ andˇ that are equal to each other. (This can be proven more formally using
mathematical induction.) ut
Theorem 12.1.8. Given any line segment and any natural number n, the line segment can be divided intonequal parts using only a straightedge and compass.