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CHAPTER 4
Circles and Regular
Polygons
Circles and regular polygons are the subject of Books III and IV of The Elements.
Euclid's abstract exposition of the interrelation of chords, arcs, and tangents lines is
augmented with the computation of the circle's circumference and area.
1. The Neutral Geometry of the Circle
Equal circles are circles that have equal radii. A chord of a circle is a line segment that
joins two of its points. A diameter is a chord that contains the center of the circle. An
arc of a circle is a portion of the circle that joins two of its points. Every chord
determines two arcs of the circle. Consequently, it takes at least three letters to denote an
arc unambiguously and the two arcs of the circle of Figure 4.1 with endpoints A and B
should be denoted, properly speaking, by arc( AEB) and arc( AFB ). Nevertheless, it is
customary to label both of these arcs arc( AB ) and to rely on the context for clarification.
A segment of a circle is the portion between a chord and either of its arcs. A sector of a
circle is the portion between two radii. The arcs determined by a diameter are each called
a semicircle. That the two semicircles determined by a diameter are equal (in length) is a
proposition that Euclid mentions in Definition 17 (Chapter 2). This observation is proved
as part of Proposition 4.1.1 below.
Figure 4.
A central angle of a circle is one both of whose sides are radii. Every arc
subtends a central angle that is either greater or less than 180
o
according as the arc is
greater or less than a semicircle. Every chord subtends a central angle that is at most
o
The following four propositions of Euclid's are established here with a single
unified proof.
PROPOSITION 4.1.1 (III.26). In equal circles equal central angles stand on equal
arcs.
PROPOSITION 4.1.1 (III.27). In equal circles central angles standing on equal
arcs are equal to one another.
PROPOSITION 4.1.1 (III.28). In equal circles equal chords cut off equal arcs, the
greater equal to the greater and the less to the less.
PROPOSITION 4.1.1 (III.29). In equal circles, equal arcs are subtended by equal
chords.
4.1 THE NEUTRAL GEOMETRY OF THE CIRCLE
3 => 1: Since ∠ AEB = ∠ A'E'B' it is possible to apply the first circle to the
second so that E falls on E' and these angles coincide. Since the circles have equal
radii it follows that arc AB falls on arc A'B'. Consequently these arcs have equal
lengths.
Q.E.D.
COROLLARY 4.1.2. In a circle all the semicircles are equal to each other.
See Exercise 1.
PROPOSITION 4.1.3 (III.3). In a circle, a radius bisects a chord not through the
center if and only if the radius and the chord are perpendicular to each other.
See Exercise 2.
EXERCISES 4.1A
- Prove Corollary 4.1.2.
- Prove Proposition 4.1.3.
- Prove that in a circle, a diameter is greater than any chord which is not a diameter.
- Prove that two chords of a circle are equal if and only if they are at equal distances from its center.
(Exercise 2.3N.2 can be used to produce a neutral proof.)
- Prove that a circle cannot contain three collinear points (III.2)
- Prove that in a circle, the radius perpendicular to a chord bisects that chord's central angle and arc.
- Prove that in a circle two equal intersecting chords cut each other into respectively equal
segments.
- Prove that of two unequal chords in a circle, the greater one is closer to the center. (This is
Proposition III.15. It can be easily proved on the basis of the Theorem of Pythagoras, but such a
proof is would not be neutral. Euclid's neutral proof is based on Proposition I.24 (Exercise
2.3Q.5).)
4.1 THE NEUTRAL GEOMETRY OF THE CIRCLE
- Let A be a given point and p a circle centered at C. If the point P moves along the circle p
prove that the midpoint of AP describes a circle centered at the midpoint of CA. (See Exercises
3.1D.7-8. It is necessary to consider three cases, depending on the relative positions of A and p. )
- Construct the midpoint of a given arc on a given circle.
- Given an arc of a circle, construct the center of the circle.
- Given points A, B, C, D construct a circle through A and B whose center is equidistant from C
and D.
- Given a point A inside a circle, construct a chord that is bisected by A. Prove that this chord is
the shortest of all the chords through A.
- Given an angle α and a line segment a, construct a circle whose center is on one side of α and
which cuts a segment equal to a on the other side.
- Given a circle p and a point A outside it, construct a straight line through A which cuts the
circle so that the segment from A to the circle equals the segment in the circle. (See Exercises
3.1.D7-8.)
- Comment on Proposition 4.1.1 in the context of the following geometries:
a) spherical; b) hyperbolic; c) taxicab; d) maxi.
- Comment on Proposition 4.1.3 in the context of the following geometries:
a) spherical; b) hyperbolic; c) taxicab; d) maxi.
An infinitely extended straight line is said to be tangent to a circle if they have
exactly one point in common, and that point is called their point of contact.
PROPOSITION 4.1.4 (III.16, 18). If a straight line intersects a circle, then they are
tangent if and only if the straight line is perpendicular to the radius through the point of
contact.
GIVEN: Circle ( C; CP ), straight line PT (Fig. 4.3).
TO PROVE: PT is tangent to ( C; CP ) if and only if CP
⊥ PT.
4.1 THE NEUTRAL GEOMETRY OF THE CIRCLE
- Prove that if two circles are internally tangent then the line joining their centers contains the point
of contact.
- Prove that if two circles are tangent to each other then they have a common tangent line at their
point of contact.
- Prove that if two circles lie outside each other then they have four different common tangent lines.
- Let m and n be common tangents to unequal circles such that both circles lie inside one of the
angles formed by these tangents. Prove that the line joining the centers of the circles bisects this
angle.
- Let m and n be common tangents to unequal circles such that the circles lie in vertically
opposite angles formed by these tangents. Prove that the line joining the centers of the circles
bisects these angles.
- Given two circles with the same center and unequal radii, prove that all the chords of the larger
circle that are tangent to the smaller circle have the same length.
- Construct a circle with a given radius tangent to a given line.
- Construct a circle with a given radius, tangent to a given line, and containing a given point.
- Construct a circle containing a given point and tangent to a given straight line at a given point on
the line.
- Construct a circle that is tangent to two given parallel straight lines.
- Construct a circle that is tangent to two given parallel straight lines and contains a given point
between them. How many solutions are there?
- Construct a circle that is tangent to two intersecting straight lines.
- Construct a circle that is tangent to two given intersecting straight lines and contains a given point.
- Construct a circle that is tangent to two given parallel straight lines as well as to a given third line
that intersects them.
- Given a circle p and a point A construct a straight line containing A such that its segment inside
p has a given length. (Hint: See Exercise 11.)
- Construct a point such that the lengths of the tangents from it to two given circles are given.
- Comment on Proposition 4.1.4 in the context of the following geometries:
a) spherical; b) hyperbolic; c) taxicab; d) maxi.
The following proposition was proved by Euclid in its entirety. The proof offered
in this text is incomplete in two ways. In the first place, the argument is restricted to
rational values of the ratios in question. Moreover, given an angle ∠ ABC and a positive
4.1 THE NEUTRAL GEOMETRY OF THE CIRCLE
integer m , this argument makes use of the angle
(∠ ABC ) / n even though it has not been
demonstrated that such an angle can be constructed within Euclid's system.
PROPOSITION 4.1.5 (VI.33). In equal circles, central angles are proportional to
the arcs on which they stand.
GIVEN: Equal circles with centers G and H respectively (Fig. 4.4).
TO PROVE:
∠ BGL
∠ EHN
arc ( BL )
arc ( EN )
Figure 4.
SUPPORTING ARGUMENT: The argument is limited to the case where the given
ratios are rational. In other words, it is assumed that there exist positive integers m and
n such that
∠ BGL
∠ EHN
m
n
i.e.,
∠ BGL
m
∠ EHN
n
Let α be an angle such that
α =
∠ BGL
m
∠ EHN
n
4.1 THE NEUTRAL GEOMETRY OF THE CIRCLE
circumference of earth
distance from Alexandria to Syene
o
0
Figure 4.
from which he concluded that the circumference is 50⋅5000 = 250,000 stadia. In order
to make his answer divisible by 60 (probably because of the influence of the Babylonian
sexagesimal number system) he adjusted this result to 252,000 stadia. The standard
stade of the time had a length of 178.6 meters which converts his rounded estimate to
45,007 km, an overestimate of 12.3%, since the circumference of the earth is actually
40,075 km.
EXERCISES 4.1C
- A circle has circumference 10 ft. Find the lengths of the arcs that subtend the following angles at
the center of the circle:
a) 10
o
b) 30
o
c) 90
0
d) 110
o
e) 120
o
f) 180
o
- A location on earth has latitude 25
o
N. Find its distance from the equator and from the North
Pole.
- A location on earth has latitude 70
o
N. Find its distance from the equator and from the North
Pole.
4.1 THE NEUTRAL GEOMETRY OF THE CIRCLE
- A location on earth has latitude 70
o
S. Find its distance from the equator and from the North
Pole.
- A location on earth has latitude 70
o
S. Find its distance from the equator and from the North
Pole.
- A location on earth lies 2000 km north of the equator. Find its latitude.
- A location on earth lies 1234 km north of the equator. Find its latitude.
- A location on earth lies 1000 km south of the equator. Find its latitude.
- A location on earth lies 617 km south of the equator. Find its latitude.
- Comment on Proposition 4.1.5 in the context of the following geometries:
a) spherical; b) hyperbolic; c) taxicab; d) maxi.
2. The Non-Neutral Geometry of the Circle
The next proposition is one of the most surprising in The Elements. Unlike those
appearing the previous section, its implications are quite unexpected.
PROPOSITION 4.2.1 (III.20). In a circle, the angle at the center is double of the
angle at the circumference, when the angles have the same arc as base.
GIVEN: Points A, B, C on the circumference of a circle centered at E (Fig. 4.6).
TO PROVE: ∠ BEC = 2 ∠ BAC.
Figure 4.
4.2 THE NON-NEUTRAL GEOMETRY OF THE CIRCLE
Figure 4.
PROOF: See Exercise 1.
This proposition is somewhat counterintuitive. Suppose the points A and B in
Figure 4.8 are fixed whereas P slides clockwise around the circle occupying positions
P
1
, P
2
, ..., P
5
successively. Proposition 4.2.2 implies that as long as the point P
remains in the interior of the upper (or longer) arc( AB) the angle APB retains a
constant (acute) value. When P passes through A or B, APB is no longer an angle.
Finally, when P is in the interior of the shorter (or lower) arc( AB) the angle APB
Figure 4.8 A discontinuous function.
assumes a different (obtuse) value. In other words, even though the point P moves in a
continuous manner, ∠ APB varies as a discontinuous function of the position of P.
4.2 THE NON-NEUTRAL GEOMETRY OF THE CIRCLE
PROPOSITION 4.2.3 (III.31). In a circle, the angle subtended by a diameter from
any point on the circumference is a right angle.
See Exercise 2.
PROPOSITION 4.2.4 (III.32) Let AB be a chord of a circle and let
AT be any
straight line at A. Then the line
AT is tangent to the circle if and only if ∠ TAB is
equal to the angle at the circumference subtended by the intercepted arc.
GIVEN: Circle p with chord AB , straight line
AT (^) , arc( AB ) (Fig 4.9).
TO PROVE:
AT is tangent to p if and only if ∠ TAB equals the angle at the
circumference of p subtended by arc( AB ).
PROOF: Let AD be the diameter of the circle containing A , and join BD (Figure 4.9).
Figure 4.
By Proposition 4.2.3 ∠ ABD = 90
o
. Hence the following statements are all equivalent to
each other:
AT (^) is tangent to the circle
∠ DAT = 90
o
o
Q.E.D.
4.2 THE NON-NEUTRAL GEOMETRY OF THE CIRCLE
PROPOSITION 4.2.6 (III.22). The opposite angles of a cyclic quadrilateral are
equal to two right angles.
See Exercise 3.
EXERCISES 4.2A
- Prove Proposition 4.2.2.
- Prove Proposition 4.2.3.
- Prove Proposition 4.2.6.
- Prove that in a circle parallel chords enclose equal arcs.
- Prove that if the quadrilateral ABCD is cyclic, then the exterior angle at A equals the interior
angle at C.
- In a circle the extensions of the chords AB and KL intersect in a point P outside the circle.
Prove that ∠ AKP = ∠ LBP and ∠ BKP = ∠ LAP.
- In a circle the extensions of the chords AB and CD intersect in a point P outside the circle.
Prove that PA ⋅ PB = PC ⋅ PD (Proposition III.35).
- In a circle the chords AB and CD intersect in a point P inside the circle. Prove that PA ⋅ PB =
PC ⋅ PD (Proposition III.36).
- Prove that two equal and parallel chords in a circle constitute the opposite sides of a rectangle.
- Prove that if the hexagon ABCDEF is cyclic and the interior angles at A and D are equal, then
BC || EF.
- In the cyclic quadrilateral ABCD, AD = BC. Prove that the interior angles at A and B are equal
to each other (as are those at C and D ).
- Prove that the sum of the interior angles at A, C and E in the cyclic hexagon ABCDEF is four
right angles.
13*. Prove that if the perpendicular chords AB and CD of a circle intersect at the point M (inside the
circle) then the straight line through M that is perpendicular to AD bisects the chord BC.
- Prove that every cyclic rhombus is a square.
- Prove that if A and B are two distinct points and D is any other point on AB then the locus of
all the points P in the plane such that
AP
PB
=
AD
DB
is a circle. (This is the circle of Apollonius.)
- State and prove the converse of Proposition 4.2.6.
4.2 THE NON-NEUTRAL GEOMETRY OF THE CIRCLE
- Given a line segment AB , construct the circle which consists of all the points from which AB
subtends an angle of 90
o
- Given a line segment AB , construct the arc which consists of all the points from which AB
subtends an angle of 60
o
- Given a line segment AB , construct the arc which consists of all the points from which AB
subtends an angle of 120
o
- Given a line segment AB , construct the arc which consists of all the points from which AB
subtends an angle equal to a given angle α.
- Construct a triangle given the data:
a) a, h b
, h c
b) a, h a
, α c) a, m a
, α
d) a + b + c, h a
, α.
- Construct a parallelogram given its two diagonals and one of its angles.
- Given line segment AB and CD and angles α and β , construct a point P such that ∠ APB =
α and ∠ CPD = β.
- In a given Δ ABC construct a point P such that ∠ APB = ∠ BPC = ∠ CPA.
- Comment on Proposition 4.2.1 in the context of the following geometries:
a) spherical; b) hyperbolic; c) taxicab; d) maxi.
- Comment on Proposition 4.2.2 in the context of the following geometries:
a) spherical; b) hyperbolic; c) taxicab; d) maxi.
- Comment on Proposition 4.2.3 in the context of the following geometries:
a) spherical; b) hyperbolic; c) taxicab; d) maxi.
- Comment on Proposition 4.2.6 in the context of the following geometries:
a) spherical; b) hyperbolic; c) taxicab; d) maxi.
29(C). Use a computer application to verify the following propositions: a) 4.2.1 b) 4.2.
c) 4.2.3b d) 4.2.4.
Three (or more) straight lines are said to be concurrent if they all contain the
same point.
PROPOSITION 4.2.7. The three perpendicular bisectors of the sides of a triangle
are concurrent.
4.2 THE NON-NEUTRAL GEOMETRY OF THE CIRCLE
PROPOSITION 4.2.9. The bisectors of the three interior angles of a triangle are
concurrent.
GIVEN: Δ ABC, AA', BB', CC' are the bisectors of ∠ BAC, ∠ ACB, ∠ ABC,
respectively (Fig. 4.12).
TO PROVE: AA', BB', CC' are concurrent.
Figure 4.
PROOF: Since
1
2
1
2
o
[PN 2.3.21]
it follows from Postulate 5 that BB' and CC' intersect in some point D. Let E, F and
G be those points on AB, BC, CA respectively such that DE
⊥ AB, DF
⊥ (^) BC, and
DG
⊥ (^) AC. Then
DE = DF = DG [PN 2.3.31]
∴ DE = DG [CN 1]
∴ DA bisects ∠ BAC [PN 2.3.32]
Q.E.D.
4.2 THE NON-NEUTRAL GEOMETRY OF THE CIRCLE
A circle that lies in the interior of a triangle and is tangent to all of its sides is said
to be inscribed in the triangle. Its center and radius are, respectively, the triangle’s
incenter and inradius.
PROPOSITION 4.2.10 (IV.4). In a given triangle to inscribe a circle.
See Exercise 2.
EXERCISES 4.2B
1. Prove Proposition 4.2.8.
- Prove that similar triangles have circumradii that are proportional to their sides.
3. Prove Proposition 4.2.10.
- Prove that similar triangles have inradii that are proportional to their sides.
- Prove that the circumcenter of a right triangle is the midpoint of its hypotenuse.
- Prove that if the circumcenter of Δ ABC lies inside the triangle then the triangle is acute, if the
center is on a side the triangle is right, and if the center is outside the triangle then the triangle is
obtuse.
- Prove that the circumcenter of an acute triangle lies inside the triangle.
- Prove that the circumcenter of an obtuse triangle lies outside it.
- Prove that the area of the triangle equals the product of half its perimeter with the inradius. (Hint:
Examine the three triangles formed by the center of the circle with the triangle's three sides.)
- In a given circle inscribe a triangle similar to a given triangle.
- Prove that the altitudes of the triangle are concurrent. (Hint: Through each vertex of the triangle
draw a line parallel to the opposite side. Then show that the altitudes in question are the
perpendicular bisectors of the triangle formed by these parallels.)
- Comment on Proposition 4.2.7 in the context of the following geometries:
a) spherical; b) hyperbolic; c) taxicab; d) maxi.
- Comment on Proposition 4.2.8 in the context of the following geometries:
a) spherical; b) hyperbolic; c) taxicab; d) maxi.
- Comment on Proposition 4.2.9 in the context of the following geometries:
a) spherical; b) hyperbolic; c) taxicab; d) maxi.
- Comment on Proposition 4.2.10 in the context of the following geometries:
a) spherical; b) hyperbolic; c) taxicab; d) maxi.
