Take K

Proposition 29.-Theorem.

In equal circles equal arcs are subtended by equal chords. Let ABC, DEF be equal circles, and let BGC, EHF be equal arcs in them, and join BC, EF.

The chord BC shall be equal to the chord EF.

CONSTRUCTION.-Take K, L, the centres of the circles

and L the (III. 1), and join BK, KC, EL, LF.

centres.

K

Then

4 BKCELF,

and so base BC= base

EF.

In the trianglesADC and CDB,

PROOF. Because the arc BGC is equal to the arc EHF (Hyp.), the angle BKC is equal to the angle ELF (III. 27).

And because the circles ABC, DEF are equal (Hyp.), their radii are equal (III. def. 1).

Therefore the two sides BK, KC are equal to the two sides EL, LF, each to each; and they contain equal angles; Therefore the base BC is equal to the base EF (I. 4). Therefore, in equal circles, &c. Q.E.D.

Proposition 30.-Problem.

To bisect a given arc, that is, to divide it into two equal parts.

Let ADB be the given arc.

It is required to bisect it.

CONSTRUCTION.-Join AB, and bisect it in C (I. 10). From the point C draw CD at right angles to AB (I. 11), and join AD and DB.

Then the arc ABD shall be bisected in the point D.

PROOF.-Because AC is equal to CB (Const.), and CD is common to the two triangles ACD, BCD;

The two sides AC, CD are equal to the two sides BC, CD, each to each;

And the angle ACD is equal to the angle BCD, because each of them is a right angle (Const.);

Therefore the base AD is equal to the base BD (I. 4).
But equal chords cut off equal arcs, the greater equal to
the greater, and the less equal to the less (III. 28);

And each of the arcs AD, DB is less than a semicircle,
because DC, if produced, is a diameter (III. 1, cor.);
Therefore the arc AD is equal to the arc DB.
Therefore, the given arc is bisected in D.

Proposition 31.-Theorem.

Q.E.F.

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.

Let ABC be a circle, of which BC is a diameter, and E the centre; and draw CA, dividing the circle into the segments ABC, ADC, and join BA, AD, DC.

The angle in the semicircle BAC shall be a right angle; The angle in the segment ABC, which is greater than a semicircle, shall be less than a right angle;

The angle in the segment ADC, which is less than a semicircle, shall be greater than a right angle.

CONSTRUCTION.-Join AE, and produce BA to F.
PROOF.-Because EA is equal to EB (I. Def. 15),
The angle EAB is equal to the angle

EBA (I. 5);

And, because EA is equal to EC,
The angle EAC is equal to the angle
ECA;

Therefore the whole angle BAC is equal to the two angles ABC, ACB (Ax. 2).

But FAC, the exterior angle of the triangle ABC, is equal to the two angles ABC, ACB (I. 32).

F

base AD=

base BD.

< BAE + < EAC,

< BAC=

< ABC +

< ACB=

4 FAC and.. a

Therefore the angle BAC is equal to the angle FAC right angle.

(Ax. 1),

.". 4 ABC <a right angle.

Hence
< ADC >
a right

angle, by
Prop. 32.

And therefore each of them is a right angle (I. Def. 10); Therefore the angle in a semicircle BAC is a right angle. And because the two angles ABC, BAC, of the triangle ABC, are together less than two right angles (I. 17), and that BAC has been shown to be a right angle;

Therefore the angle ABC is less than a right angle.

Therefore the angle in a segment ABC, greater than a semicircle, is less than a right angle.

And, because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are together equal to two right angles (III. 22);

Therefore the angles ABC, ADC are together equal to two right angles.

But the angle ABC has been shown to be less than a right angle;

Therefore the angle ADC is greater than a right angle; Therefore the angle in a segment ADC, less than a semicircle, is greater than a right angle.

Therefore, the angle, &c. Q.E.D.

COROLLARY.-From this demonstration it is manifest that, if one angle of a triangle be equal to the other two, it is a right angle.

For the angle adjacent to it is equal to the samo two angles (I. 32).

And, when the adjacent angles are equal, they are right angles (I. def. 10).

Proposition 32.-Theorem.

The angles contained by a tangent to a circle and a chord drawn from the point of contact are equal to the angles in the alternate segments of the circle.

Let EF be a tangent to the circle ABCD, and BD a chord drawn from the point of contact B, cutting the circle.

The angles which BD makes with the tangent EF shall be equal to the angles in the alternate segments of the circle; That is, the angle DBF shall be equal to the angle in the segment BAD, and the angle DBE shall be equal to the angle in the segment BCD.

CONSTRUCTION.-From the point B draw BA at right

angles to EF (I. 11).

Take any point C in the circumference BD, and join AD, DC, CB.

PROOF.-Because the straight line EF touches the circle ABCD at the point B (Hyp.), and BA is drawn at right angles to the tangent from the point of contact B (Const.),

The centre of the circle is in BA (III. E

19).

The centre

is in BA.

Therefore the angle ADB, being in a semicircle, is a right.. ADB is angle (III. 31).

Therefore the other two angles BAD, ABD are equal to right angle (I. 32).

But ABF is also a right angle (Const.);

a right angle, a and

Therefore the angle ABF is equal to the angles BAD, ABD. From each of these equals take away the common angle ABD; Therefore the remaining angle DBF is equal to the remaining angle BAD, which is in the alternate segment of the circle (Ax. 3).

< BAD + < ABD = a right angle

=

ABF.

And because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are together equal to two Also, right angles (III. 22).

< BCD + < BAD=

But the angles DBF, DBE are together equal to two 2 right right angles (I. 13);

Therefore the angles DBF, DBE are together equal to the angles BAD, BCD.

angles

=

And the angle DBF has been shown equal to the angle DBF BAD;

Therefore the remaining angle DBE is equal to the angle BCD, which is in the alternate segment of the circle (Ax. 3). Therefore, the angles, &c. Q.E.D.

Proposition 33.-Problem.

Upon a given straight line to describe a segment of a circle, containing an angle equal to a given rectilineal angle.

Let AB be the given straight line, and C the given rectilineal angle.

+ DBE.

... 4 DBE = 4 BCD.

Angle in a

It is required to describe, on the given straight line AB, a segment of a circle, containing an angle equal to the angleC. CASE I.-Let the angle C be a right

H

A

F

B

angle.

CONSTRUCTION.-Bisect AB in F

(I. 10).

From the centre F, at the distance FB, describe the semicircle AHB.

Then AHB shall be the segment required.

PROOF. Because AHB is a semicircle, the angle AHB semicircle in it is a right angle, and therefore equal to the angle C (III. 31).

is a right

angle.

At point A

CASE II.-Let C be not a right angle.

CONSTRUCTION.-At the point A, in the straight line AB, BADC; make the angle BAD equal to the angle C (I. 23).

make

and draw

AEat right angles to AD.

From F,

From the point A draw AE at right angles to AD (I. 11).
Bisect AB in F (I. 10).

From the point F draw FG at right angles to AB (I. 11),

middle of and join GB.

AB, draw perpendicular, meeting

AE in G.

Because AF is equal to BF (Const.), and FG is common to the two triangles AFG, BFG;

The two sides AF, FG are equal to the two sides BF, FG, each to each;

And the angle AFG is equal to the angle BFG (Const.);
Therefore the base AG is equal to the base BG (I. 4).
And the circle described from the centre G, at the dis-
tance GA, will therefore pass through the point B.

Let this circle be described; and let it be AHB.

The segment AIIB shall contain an angle equal to the given rectilineal angle C.