WE CAN MAKE all the constructions necessary to this geometry, or to any other elementary geometry, by means of three instruments: a ruler marked in inches, centimeters, or the like; a protractor; and a pair of com-passes. The marked ruler, often called a scale, and the protractor embody our fundamental assumptions about numbering distances and angles as set down in Principles 1 and 3. We need the compasses in order to draw circles.
Architects and draftsmen, who are continually making geo-metric constructions, find these instruments indispensable.
From the time of the ancient Greeks down to the present, geometers have been fascinated by the question of what constructions are possible with only two very simple instru-ments: the straightedge, or unmarked ruler, and compasses.
This question is interesting also because of its intimate rela-tion to algebra. Consider, for example, any geometric construction in which we attempt to construct a desired length, x, from its relation to given lengths, a, b, C, • • • ,
using only a straightedge and compasses. Whether or not we can do so depends upon the algebraic relationship of xto a, b,C, ••••We can construct the desired length x with these two instruments whenever the relationship of x to 165
a, b,C, • • • involves ultimately only addition, subtraction, multiplication, division, and the extraction of square roots. In all other cases it is impossible to construct x with the use of only a straightedge and compasses.
Thus with these two instruments we can construct a length x equal to when aand bare known, but we cannot construct a length x equal In general, it is impossible to divide a given angle into a given number of equal parts by means of straightedge and compasses alone. This constitutes the chief difference between constructions that are possible with these two instruments and constructions that are possible with scale, protractor, and compasses.
First we shall consider constructions that are done by means of the three instruments, scale, protractor, and com-passes. Then we shall see how many of these are possible when we use only the straightedge and compasses. Prob-ably you are already familiar with some of these construc-tions. Now, however, you can prove the correctness of such constructions.
By means of the marked ruler or scale we can draw a line through two points, measure a line segment, layoff a line of given length, and divide a line segment intonequal parts. Thus,ifthe line segmentABis to be divided into six equal parts, we can use any scale we like to measure its length, then take one-sixth of this length and lay it off five times from A or B. Fig. 1 shows two ways of doing this.
I
The numbers corresponding to A and B may be given to us, instead of being found by means of a scale. Thus in Fig. 2 on the next page we are given the straight line
with the point A corresponding to and the point B corresponding to If we have to divide this line seg-ment AB into six equal parts, we must find the numbers corresponding to the five points of division. If these numbers are to be found correct to four figures, we must express and with four-figure accuracy before we subtract and divide by 6.
317 4.269 5.395 6.521 7.647 8.773
I I I I I I I
A B
Fig. 2
Similarly, ifABis to be divided into three parts propor-tional to the given lengths 1, m, and n, we measure the given lengths with the scale and layoff lengths equal to the following fractional parts ofAB:
1 m n
l+m+n' l+m+n' and l+m+n'
On a sheet of paper layoff lines equal in length to AB, 1, m, and n, as given in Fig. 3. Then divide AB into three parts proportional to1, m, and n,
A B
m n
Fig. 3
Fig. 4 s
r
Another construction fre- q quently needed in geometry is to layoff the length that is the fourth proportional to three given lengths. That
is, given the three lengths q, r, and s in Fig. 4, we must layoff the length t such that After we have
meas-r t
ured the lengths q,r, and s, we can calculate the value of t from the equation t=rs and then layoff this length on
q any desired line.
Find the length of the fourth proportional to q, r, and s in Fig. 4 on page 167.
The principles underlying the constructions which have just been discussed can also be applied to angles by means of the protractor. For example, we can divide a given angle into seven equal parts, divide it also into n parts proportional to n given angles, and find the fourth propor-tional to three given angles.
PERPENDICULARS. To draw the perpendicular to a given line at a given point of the line, we need only the pro-tractor. When the given pointP isnot on the given line AB, however, we may use the construction employed in proving Principle 11. See Fig. 5.
To draw a line from P perpendicular toAB,first join P to any point Q of AB. Construct QR so that LBQR equals L BQP. On QR layoff QP' equal to QP. Then pp'will be perpendicular toAB.
p
Fig. 6
PARALLELS. To draw the line through P parallel toAB, first draw the perpendicular throughP toABby the previ-ous construction, and then by means of a protractor draw the perpendicular to this perpendicular at P. See Fig. 6.
EXERCISES
In making the constructions called for in the following exercises, do not mark any lines or other figures printed in the book. First copy the given figure and then make the construction on your own copy of the figure. Use only scale and protractor in these exercises. In each case you will have to decide whether to use an inch-scale or a centimeter-scale.
1. Divide the line segment CD in Fig. 7 into five equal parts. Use as finely divided a scale as you can obtain to help you mark off the required points as accurately as possible.
c o
Fig. 7
2. Divide line segmentEF (Fig. 8) into six equal parts.
Fig. 8
F
H Fig. 9
3. What numbers should be assigned to the points which divide GH in Fig. 9 into four equal parts?
m Fig. 10 4. Divide the line segment CD
in Fig. 7 into two parts proportional to the two line segments 1and m in Fig. 10.
5. Divide EF in Fig. 8 into three parts proportional to the line segmentsr,S, andt in Fig. 11.
r s t
Fig. 11
169
6. Find the fourth proportional to the three lengths
T, s, and tin Fig. 11 on page 169.
7. Find the fourth proportional to the three lengths t,s, and rin Fig. 11.
8. Divide angleAOB (Fig. 12) into three equal parts.
Fig. 12
o
0Fig. 13
Fig. 14 9. Divide the angle AOB into parts proportional to the angles CODandEOF shown in Fig. 13.
10. Find the fourth proportional .p to the anglesAOB, COD,and EOF.
11. In Fig. 14 draw the line throughP perpendicular to AB.
12. In Fig. 14 draw the line through P parallel toAB.
CIRCLES. So far we have considered only constructions that require scale and protractor. Certain other construc-tions that demand the actual drawing of a circle require compassesinaddition to these two instruments. We shall not actually make such constructions at this time, how-ever, because the most important of themwillbe done later by straightedge and compasses alone; it will be evident then how we ought to proceed incase wewish touse scale, protractor, and compasses instead.
REGULAR POLYGONS. If you had to construct with scale and protractor a regular polygon of 7, or of n, sides such that each side would be equal to a given line segment
Fig. 15
AB, you would need to recall what have learned in Exercises 6 and 10, page 85, about the angles of a regu-lar polygon. But once you had discovered how large each angle of the polygon would have to be, you would have no trouble in completing the construction.
Also, if you had to inscribe a regular polygon ofn sides in a given circle, using only scale and protractor, you would need only to consider the relation between the sides of the polygon and angles at the center of the circle in order to figure out how to make the construction.
It is clear that any construction involving the laying off of lengths and angles can be made with scale and protractor, and to any desired degree of accuracy. Thus we can make the above two constructions concerning regular polygons as accurately as we wish by means of scale and protractor. But we cannot make these two constructions with straightedge and compasses except in certain special cases. For each of these constructions in-volves the division of 1800 or 3600 inton equal parts, and ordinarily we cannot divide an angle into n equal parts with straightedge and compasses. stated previously, this constitutes the chief difference between constructions that are possible with scale, protractor, and compasses and those that are possible with only straightedge and compasses.
EXERCISES
1. Using scale and protractor, A - - - B construct a regular octagon with
each of its sides equal to AB in Fig. 15.
2. Construct a regular polygon of nine sides with each of its sides equal to AB in Fig. 15, using scale and protractor.
3. Inscribe a regular octagon in a given circle, using scale and protractor.
4. Inscribe a regular polygon of nine sides in a given circle.
From a practical standpoint, constructions with straight-edge and compasses are neither more accurate nor less accurate than constructions with scale and protractor.
Moreover, as stated above, there are some scale and pro-tractor constructions that cannot be done by straightedge and compasses. Theoretically, however, those construc-tions that can be done by straightedge and compasses are absolutely accurate.
From this point on in this chapter the constructions that we shall consider will be done by means of straight-edge and compasses.
To construct the perpendicular bisector of a line segment.
""\,
,,
"
\,
\,
,
\,
,
,,, I
I
\
:
\\ ,/
' Fig. 17
GIVEN: Line segmentAB (Fig. 16).
TO CONSTRUCT: The perpendicular to AB at its mid-point.
ANALYSIS: Let us assume that the construction has been performed, that PQisthe required perpendicular, and
see what we can learn from the completed figure. All the points on PQ are equidistant from A and B (by Principle 10). We have, therefore, to establish two points that are equidistant from A and B in order to determine the line PQ.
CONSTRUCTION: With A as center and a radius greater than draw an arc. See Fig. 17 on page 172. WithB as center and the same radius draw a second arc.
The arcs intersect at the points R and R'. (See Ex. 13, page 142.) The line through Rand R'will be perpendicu-lar to AB at its mid-point. Proveit.
Is it possible to choose too small a radius for the arcs?
Too large a radius?
In any construction involv-ing the intersection of arcs you ought to try to have the arcs
cross at approximately right Fig. 18 angles. Ifyou draw arcs that
cross like those in Fig. 18, you cannot be sure of the exact location of the crossing. This is purely a practical mat-ter. Theoretically, one sort of intersection is as good as another.
To construct an angle equal to a given angle.
\
,.
•I\
Fig. 19
GIVEN: Angle A (Fig. 19).
TO CONSTRUCT: An angle equal to LA with vertex atA'.
Complete the construction. Prove that your construc-tion is correct.
To bisect a given angle.
A
Fig. 20
GIVEN: Angle A (Fig. 20).
TO CONSTRUCT: An angle equal to
Complete the construction and prove that it is correct.
iD
Through a given point, to draw the line that is parallel to a given line.
GIVEN: Line l and point P not on l.
TO CONSTRUCT: The line through P parallel to l.
Three methods of making the construction are discussed below.
First method: Through P draw a random linetcuttingl
atA (Fig. 21). Then construct an angle with vertex atP and with one side lying along t so that the angle is equal to, and corresponds to, one of the angles with vertex atA.
The other side of this angle establishes the parallel to l through P. Why?
Fig. 22
Second method: With any convenient point 0 as center (Fig. 22)draw a circle through P intersecting l at A and B.
With B as center and AP as radius draw an arc intersect-ing the circle atQ. PQwill be parallel to AB. Why? Sug-gestion: Show by
construct-S/ Fig. 23
To divide a given line segment into parts pro-portional to n given line segments.
s R
r Q
r
, ,,,
,, ,,,
,,,, ,,
,,,, ,,,,,
,
s 5
x
Fig. 24
GIVEN: Line segments AB, p, q,r, and s (Fig. 24).
TO CONSTRUCT: Points on AB which divide it into seg-ments proportional to p, q, r, and s.
ANALYSIS: Let us assume that the construction has been made and that H, J, K are the desired points. If now the segments p, q, r,and s are laid off in order along the line AX (drawn at random through A), then HP, JQ, KR, andBB will all be parallel. Why?
CONSTRUCTION: If therefore we layoff the given seg-ments p, q, r, and s in order along a line AX and draw BB, it will be easy to construct the points H, J, and K.
How?
Complete the construction and prove that your con-struction is correct.
EXERCISES
1. How could the construction discussed on page 176 be used to divide a given line segment into n equal parts?
Divide a line 2 inches long into 5 equal parts by this method.
2. Another construction for dividing a given line seg-ment into n equal parts is shown in Fig. 25.
R
Fig.2S
CONSTRUCTION: With any convenient radius and with A and B as centers draw two arcs intersecting in points P andQ. DrawAPand extend it to a point R such that AR is equal to (n-1)AP. Draw RQ. It intersects AB atN,making BNequal to of AB. Hence BNisone of
n
the required equal parts ofAB. Why? Suggestion: Draw BQand use trianglesBNQand ANR.
To construct the fourth proportional to three given line segments.
k m
A k K m M
X ,
,,,, ,,,
\\
,,,
y Fig. 26
GIVEN: Three line segments k, l, and m (Fig. 26).
TO CONSTRUCT: Line segmentn such
ANALYSIS: Let us assume that n has been constructed.
If then k and m are laid off in order along the line AX, and land n are laid off in order along AY, MN will be parallel to KL. Why?
CONSTRUCTION: Draw any two intersecting lines AX and AY. On AXlayoff lengthsk and m, and on AY lay off lengthl.
Complete the construction and prove that itis correct.
To construct the perpendicular to a given line at a given point of the line.
GIVEN: Linel and point P onl.
TO CONSTRUCT: The perpendicular tol atP.
CONSTRUCTION: Fig. 27 suggests the method to use.
Make the construction and prove that it iscorrect.
9
Q...
R P
Fig. 27
s
/
•...•...-,A P
Fig. 28
•
The constructions shown in Figs. 28, 29, and 30 are sometimes used in constructing a perpendicular to l at a point P very near the edge of the paper. (In Fig. 30 point 0 is taken at random, and a circle with radius OP is drawn.) Such constructions have little practical value, but it is interesting to see why each of them is correct.
We shall do this in Ex. 5 on page 181.
Fig. 31 Fig. 30 ...
.. Q Q
v
t
Q
A P
Fig. 29
To construct through a given point the perpendic-ular to a given line.
GIVEN: Linel and pointP.
TO CONSTRUCT: The perpen-dicular to l through P.
CONSTRUCTION: Fig. 31 sug-gests the method to use. Com-plete the construction and prove that it is correct.
Which is easier, to make this construction with straight-edge and compasses or with scale and protractor?
Show that when P lies on l this construction can still be performed, and that it is equivalent to the construc-tion shown in Fig. 27 on page 179.
To construct the mean proportional between two given line segments.
GIVEN: Line segments a and b (Fig. 32).
TO CONSTRUCT: A line segmentmsuch
ANALYSIS: The construction of the fourth proportional to m three given line segments gives us no hint here. We have met a mean proportional before in b Ex. 14, page 85, and in Ex. 26:
Fig. 32 page 150, however, and may get a suggestion from these sources.
Complete the construction and prove that it is correct.
The length of the required mean proportional m satis-the equation m2=ab or m= In particular, if b= 1, m= We may use this method, therefore, to construct a line segment of length equal to or the like, provided we know also the unit length.
EXERCISES
E Fig. 33
, ,
1. Given an arc of a circle, find C
its mid-point.
2. Given the rectangle ABCD and pointE on AB (Fig. 33), con-struct a rectangle similar to ABCD and with one side equal to AE.
Fig. 34
GIVEN: Points A, B, C, not on the same straight line. See Fig. 34.
3. Given the unit length, construct line segments of lengths and
4. Construct the fourth proportional to three line seg-ments of lengths 5, and 3
5. Prove the correctness of the constructions shown in Figs. 28-30, page 179.
To construct the circle through three given points.
TO CONSTRUCT: The circle passing throughA, B, and C.*
ANALYSIS: Since the center of the required circle must be equidistant from A, B, and C, we have to locate a point satisfying this condition. We have seen in Theorem 10 that allpoints equidistant from the
ends of a line segment lie on the perpendicular bisector of the line segment. Therefore the center of the circle through A, B, and C must lie on the perpendicular bi-sectors of AB, BC, and AC. Show that these three per-pendicular bisectors meet in a point. This can best be done by showing first that two of these meet in a point, and then that their point of intersection lies on the third.
CONSTRUCTION: Fig. 34 suggests the method to use.
The work involved in constructing the perpendicular bi-sectors can be shortened in this case by keeping the same radius throughout, as shown in Fig. 34. Make the con-struction and prove that it is correct.
*That there is such a circle has been proved in Ex. 1 on page 154.
Fig. 35 C
To circumscribe a circle about a given triangle.
This construction is equivalent to the preceding one.
Given an arc of a circle, to find its center.
GIVEN: An arc of a circle (Fig. 35).
TO CONSTRUCT: The center of the circle.
ANALYSIS: The center, when found, will be equidistant from all points of the arc, in particular from the three points A, B, and Cchosen at random. Where are all the points that are equidistant fromA andB? From BandC?
Make the construction and prove that it is correct.
To inscribe a circle in a given triangle.
C
GIVEN: Triangle ABC (Fig. 36).
TO CONSTRUCT: A circle tangent to the three sides of triangle ABC.
*
ANALYSIS: Assume the construction performed. Then the center 0 must be equidistant from AB, BC, and CA.
From this it follows that in the right triangles ADO and AFO, LDAO= LFAO. Why?
*That there is such a circle can be proved by showing (1) that the bisectors of any two angles of a triangle must intersect, and (2) that this point of intersection is equally distant from the three sides of the triangle and so can serve as center of the inscribed circle. We shall return to this idea in Ex. 19 on page 256.
Fig. 39
That is, 0 lies on the bisector of angleBAC. Similarly
o
lies on the bisectors of anglesB and C.CONSTRUCTION: Draw the bisectors of the three angles, A, B, and C. Theywill meet at the center of the inscribed circle.
How would you construct a circle tangent to one side of a
triangle and also tangent to the other two sides extended, as in Fig. 37?
triangle and also tangent to the other two sides extended, as in Fig. 37?