PROP. 30.-Prob.

To bisect a given arc of a circle, i. e., to divide it into two equal parts.

24,
I.
3 28, III.

D

BC, CD com.

= BCD, A

B

с

and ACD =
.. base AD = base BD.

But eq. st. lines cut off eq. arcs, the gr. = the

gr., and the less = the less;

DC passes through the cen., arcs

4 Cor. 1, III.

and

AD, D B,

each < a semicircle.

5 Conc.

.. arc AD

arc D B, and ADB is bis. in D.

Q. E. F.

COR.-Hence, by successive bisections, as in Sch. 1 and 3, Prop. 9, I., a given arc may be divided into any number of equal parts that are the powers of two, as 4, 8, 16, 32, &c.

SCH.-1. The bisection of a given rectil. angle, ADB,Prop. 9, I., implies

a bisection of AD B, an arc of a circle; but, as by Plane Geometry, a rectil. angle, except in the case of a right angle, DCB, cannot be divided into 3, 5. 6, &c., equal parts, so an arc of a circle, except in the case of a quadrant, DB, which is the circular measure of a right angle, DCB, cannot be cut into 3, 5, 6, &c., equal parts.

D

A C

B

N.B. If, as is mentioned below, other plane curves, besides the circle, had been admttted by EUCLID, any angle could also be divided into 3, 5, 6, &c., equal parts.

2. To trisect. a quadrant, AB, i. e. to divide a rt. L, ACB, into three equal parts.

C. 1 1, I.

2 9, I.

3 Sol.

On CB const. an equil. A BEC,
and bis. its BCE by CD,
the rt. ACB will be trisected,
i. e. [BCD = /DCE = / ECA.
BCE of a rt., and is,
bis. by CD;
../ BCD:

D.1 Cor.32,I. C2..

2 Conc.

=

ZDCE.

3 H. & D. 1. Also. A CB is a rt. 4, and BCE C

of a rt.

../ ACE =

of art.

;

B

..ZACE = / ECD = / DCB, i. e. the quadrant is trisected.

Or,-From A and B, with the rad. of the circle, describe arcs cutting the quadrant in D and E; join EC, DC, and the quadrant is trisected.

3. By successive bisections of the one-third of a rt., the 1-6th, 1-12th, 1-24th, &c., of a rt. ▲, or of a quadrant is obtained.

E

B

4. The division of a quadrant into five equal parts depends on P. 10, IV. 5. Since EUCLID confines himself to straight lines and circles, the trisection of an angle, or, of an arc cannot be effected by his Geometry; if, however, other curves, formed by the sections of the Cone, were admitted among the curves of our Plane Geometry, the problem could readily be solved. For, "if with twothirds of any given line, A, as a major axis, an hyperbola be described whose asymptotes," or incoincident lines, "make an angle of 120°; and if with A as a base, and a point on the branch of the hyperbola adjacent to the single third of A as a vertex, a triangle be described, the larger of the angles adjacent to A will always be double of the smaller. Consequently, one of the external angles will be triple of one of its internal and opposite angles; so that by describing on a straight line A a segment of a circle containing the supplement of any given angle less than 180°, that circle will cut the branch of the hyperbola in a point which, being joined with the further extremity of A, will give an angle equal to the given angle."-PENNY CYCL., XXV., p. 260.

6. What is required for the trisection of an arc or of an angle is the solution of the following problem ; "from a given point, as A, in the circumterence of a ABD, to draw a st. line, A XY, such that the part XY, between the circumference and a given diameter BD produced, shall be equal to the radius CA."

32, I.

B

A

H. 6, I... CX = XY,.. / XCY = / XYC;
but AXC or CAX = 2/ XCD,
and ACB / XCD + 2 ÁXC;
LACB=3 / XCD, .. arc AB 3 arc XD.
Thus XCD = of ACB, and .. arc XD =

7. One of the Trochoidal curves, known by the name of the trisectriz, is peculiarly possessed of the property of dividing any arc into three equal parts.

USE AND APP.-By this problem, the semicircle is divided into quadrants, and the quadrant into arcs of 45°, 224°, &c.; and the Mariner's Compass, as in Use 4, 9, I., into 32 equal parts called Rhumbs; but the division into single degrees cannot be performed by Euclid's Geometry.

PROP. 31.-THEOR.

In a circle, the angle in a semicircle is a right angle, but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.

CON. 10, I. Psts. 1 & 2.

DEM. Def. 15, I. 5, I. Axs. 1, 2. Def. 10, I.
22, III.
32, I. If a side of a ▲ be produced, the ext. : the two int. and
opp.s; and the three int. /s of every ▲ are together equal to
two rt. s.

17, I. Any two angles of a ▲ shall together be less than two rt. angles.

E.11 Hyp. 1.

2

Let ABC be a O, BC its diam.
and E its cen.

D

ADC, of

2. and from C let CA divide the
into segs. ABC,

which seg. ABC is a semic.
and ADC < a semic.

3 Conc. 1. Then BAC in the semic. is a

433

C.1 10, I.

2.

rt. Li

Z ABC in seg. ABC is < a rt. ▲ ;
3. and ADC in seg. ADC is > a rt. .

2 Pst. 1 & 2 3 Pst. 1.

CASE I-The L BAC in the semicircle shall be a rt. L.

D.1 Def. 15, 1.5, I.

2 Add. Ax. 2.

3 32, I.

Now, EB = EA = EC; .. ▲ EBA
= ▲ EAB, and ▲ EAC= / ECA;
add the eqs. and the whole

Zs ABC, ACB;
but in ▲ ABC, the ext.
int. Zs ABC, ACB;

=

BAC the two

FAC the two

Z FAC, and .. each is a rt. . BAC in the semic. is a rt. .

4 Ax.1.Def.10,I. .. ▲ BAC 5 Conc.

the

CASE II.-The ABC, in a seg. ABC gr. than a semic., shall be less than a rt. L.

D.1 17, I. And in ▲ ABC the s ABC, BAC are < 2 rt. / s. and that BAC is a rt.

2 Case I. 3 Conc.

Rec.

.. ABC must be a rt.

;

.

the, in a seg. > a semic., is less than a rt. 2.

CASE III.-The ADC, in a seg. ADC < a semic., shall be

COR.-If one angle of a triangle be eq. to the other two, it is a

rt. angle.

E.1 Hyp. 2 Conc.

C. Pst.

D.1 32, I.

For in ▲ ABC, let / BAC = Zs ABC + ACB; then BAC is a rt. Z.

Produce BA to F.

Z FACs ABC + ACB;

. FAC / BAC,

2 Def.10, I... Z BAC is a rt. Z.

Scu.-The converse of Prop. 31, is, "the segment which contains an acute angle is greater than a semicircle, and that which contains an obtuse angle is less than a semicircle."

2. The Demonstration which LARDNER gives of the 31st Prop. is remarkable for its elegance and brevity; it is founded on Prop. 20, Book III.;—

..the central

.. the circumferential

on a semicircle = two rt. s;

one rt.

.

Again,. the cen. on an arc less than a semic. is less than two rt.s ; .. the circumferential is < a rt. .

And the cen. ▲ on an arc gr. than a semic. is gr. than two rt. s; .. the circumferential is > a rt. .

USE AND APP.-I. From the property, Case I., P. 31, III., that the angle in a semicircle is a rt. angle, the following Problems are derived:

PROB. 1. From a point B, in a line, or at the extremity of a line, to draw a perpendicular.

From any A, out of the line, with the distance A B, describe a semic. meeting CA produced in D, and join BD; then BD is the perp.

N.B. PELITARIUS, a mathematician, often quoted by BILLINGSLEY, gives this Cor. to Pr. 31.-"If in a circle be inscribed a rectangle triangle, the side opposite unto the C right angle shall be the diameter of the circle."

PROB. 2.—From a point D, without CB a line, to draw a perp.

D

Join D, C, and bisect DC in A; and from A with AC describe a semic. and join AB; DB is the perp. required.

PROB. 3.-From a point A, without a circle, to draw a tangent.

Join A and the cen. C; bis. AC in E; and
from E with EA desc. a circle; and draw
AB and AD; then AB and AD are
both tangs. from A to the BCD.

The Demonstration may be left to the Student.

Or, join A, C; 10, I., bis. AC in D, and with DA
desc. a semicircle ABC; where the semic. cuts
the B, is the tangent point; and AB the A
tangent from A.

II. By means of a square the centre of a circle may be easily found;

For, if B the angular point of the square, CBD (as

in the fig. to Prob. 1. above) touch any point in the Oce; and if also the sides of the square, BD, BC, fall upon two other points, C, D of the Oce; then the line CD is a diam. and its middle point the centre of the circle.