Then AB shall touch the circle BCD in the point B. Because E is the center of the circles BCD, AFG, (1. def. 15.) therefore EA is equal to EF, and ED to EB;

therefore the two sides AE, EB, are equal to the two FE, ED. each to each:

and they contain the angle at E common to the two triangles AEB, FED;

therefore the base DF is equal to the base AB, (1. 4.)

and the triangle FED to the triangle AEB,

and the other angles to the other angles:

therefore the angle EBA is equal to the angle EDF:
but EDF is a right angle, (constr.)
wherefore EBA is a right angle: (ax. 1.)

and EB is drawn from the center:

but a straight line drawn from the extremity of a diameter, at right angles to it, touches the circle: (III. 16. Cor.) therefore AB touches the circle;

and it is drawn from the given point A.

Secondly, if the given point be in the circumference of the circle, as the point D,

draw DE to the center E, and DF at right angles to DE:
then DF touches the circle. (III. 16. Cor.) Q.E.F.

PROPOSITION XVIII. THEOREM.

If a straight line touch a circle, the straight line drawn from the center to the point of contact, shall be perpendicular to the line touching the circle.

Let the straight line DE touch the circle ABC' in the point C; take the center F, and draw the straight line FC. (III. 1.) Then FC shall be perpendicular to DE.

If FC be not perpendicular to DE; from the point F, if possible, let FBG be drawn perpendicular to DE.

And because FGC is a right angle,

therefore GCF is an acute angle; (I. 17.)

and to the greater angle the greater side is opposite: (1. 19.)
therefore FC is greater than FG:

but FC is equal to FB; (1. def. 15.)
therefore FB is greater than FG,

the less than the greater, which is impossible:
therefore FG is not perpendicular to DE.

In the same manner it may be shewn,

that no other line is perpendicular to DE besides FC,
that is, FC is perpendicular to DE.

Therefore, if a straight line, &c. Q. E. D.

PROPOSITION XIX. THEOREM.

If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the center of the circle shall be in that line.

Let the straight line DE touch the circle ABC in C,
and from Clet CA be drawn at right angles to DE.
Then the center of the circle shall be in CA.

For, if not, let F be the center, if possible, and join CF.
Because DE touches the circle ABC,

and FC is drawn from the center to the point of contact,
therefore FC is perpendicular to DĒ; (III. 18.)
therefore FCE is a right angle :

but ACE is also a right angle; (hyp.)

therefore the angle FCE is equal to the angle ACE, (ax. 1.)
the less to the greater, which is impossible:
therefore Fis not the center of the circle ABC.
In the same manner it may be shewn,

that no other point which is not in CA, is the center;
that is, the center of the circle is in CA.
Therefore, if a straight line, &c. Q. E. D.

PROPOSITION XX. THEOREM.

The angle at the center of a circle is double of the angle at the circumference upon the same base, that is, upon the same part of the circumference.

Let ABC be a circle, and BEC an angle at the center, and BAC an angle at the circumference, which have BC the same part of the circumference for their base.

Then the angle BEC shall be double of the angle BAC.

B

Join AE, and produce it to F.

First, let the center of the circle be within the angle BAC.
Because EA is equal to EB,

therefore the angle EBA is equal to the angle EAB; (1. 5.) therefore the angles EAB, EBA are double of the angle EAB: but the angle BEF is equal to the angles EAB, EBA; (1. 32.)

therefore also the angle BEF is double of the angle EAB: for the same reason, the angle FEC is double of the angle EAC: therefore the whole angle BEC is double of the whole angle BAC. Secondly, let the center of the circle be without the angle BAC.

It may be demonstrated, as in the first case,

that the angle FEC is double of the angle FAC,

;

and that FEB, a part of the first, is double of FAB, a part of the other therefore the remaining angle BEC is double of the remaining angle BAC.

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

PROPOSITION XXI. THEOREM.

The angles in the same segment of a circle are equal to one another.
Let ABCD be a circle,

and BAD, BED angles in the same segment BAED.
Then the angles BAD, BED shall be equal to one another.
First, let the segment BAED be greater than a semicircle.

Take F, the center of the circle ABCD, (III. 1.) and join BF, FD. Because the angle BFD is at the center, and the angle BAD at the circumference, and that they have the same part of the circumference, viz. the arc BCD for their base;

therefore the angle BFD is double of the angle BAD: (111. 20.) for the same reason the angle BFD is double of the angle BED: therefore the angle BAD is equal to the angle BED. (ax. 7.) Next, let the segment BAED be not greater than a semicircle.

Draw AF to the center, and produce it to C, and join CE.
Because AC is a diameter of the circle,

therefore the segment BADC is greater than a semicircle;
and the angles in it BAC, BEC are equal, by the first case:

for the same reason, because CBED is greater than a semicircle, the angles CAD, CED, are equal:

therefore the whole angle BAD is equal to the whole angle BED. (ax. 2.) Wherefore the angles in the same segment, &c.

PROPOSITION XXII. THEOREM.

Q. E. D.

The opposite angles of any quadrilateral figure inscribed in a circle, are together equal to two right angles.

Let ABCD be a quadrilateral figure in the circle ABCD. Then any two of its opposite angles shall together be equal to two right angles.

D

A

Join AC, BD.

And because the three angles of every triangle are equal to two right angles, (I. 32.)

the three angles of the triangle CAB, viz. the angles CAB, ABC, BCA, are equal to two right angles:

but the angle ĈAB is equal to the angle CDB, (III. 21.)

because they are in the same segment CDAB;

and the angle ACB is equal to the angle ADB, because they are in the same segment ADCB: therefore the two angles CAB, ACB are together equal to the whole angle ADC: (ax. 2.)

to each of these equals add the angle ABC;

therefore the three angles ABC, CAB, BCA are equal to the two angles ABC, ADC: (ax. 2.)

but ABC, CAB, BCA, are equal to two right angles; therefore also the angles ABC, ADC are equal to two right angles. In the same manner, the angles BAD, DCB, may be shewn to be equal to two right angles.

Therefore, the opposite angles, &c. Q. E.D.

PROPOSITION XXIII. THEOREM.

Upon the same straight line, and upon the same side of it, there cannot be two similar segments of circles, not coinciding with one another.

If it be possible, upon the same straight line AB, and upon the same side of it, let there be two similar segments of circles, ACB, ADB, not coinciding with one another.

D

B

Then, because the circumference ACB cuts the circumference ADB in the two points A, B, they cannot cut one another in any other point: (III. 10.)

therefore one of the segments must fall within the other:
let ACB fall within ADB:

draw the straight line BCD, and join CA, DA.

Because the segment ACB is similar to the segment ADB, (hyp.) and that similar segments of circles contain equal angles; (III. def. 11.) therefore the angle ACB is equal to the angle ADB,

the exterior angle to the interior, which is impossible. (1. 16.) Therefore, there cannot be two similar segments of circles upon the same side of the same line, which do not coincide. Q.E.D.

PROPOSITION XXIV. THEOREM.

Similar segments of circles upon equal straight lines, are equal to one another. Let AEB, CFD be similar segments of circles upon the equal straight lines AB, CD.

Then the segment AEB shall be equal to the segment CFD.

For if the segment AEB be applied to the segment CFD, so that the point A may be on C, and the straight line AB upon CD, then the point B shall coincide with the point D, because AB is equal to CD:

therefore, the straight line AB coinciding with CD,

the segment AEB must coincide with the segment CFD, (III. 23.) and therefore is equal to it. (I. ax. 8.)

Wherefore similar segments, &c. Q.E.D.

PROPOSITION XXV. PROBLEM.

A segment of a circle being given, to describe the circle of which it is the segment.

Let ABC be the given segment of a circle.

It is required to describe the circle of which it is the segment. Bisect AC in D, (1. 10.) and from the point D draw DB at right angles to AC, (1. 11.) and join AB.

First, let the angles ABD, BAD be equal to one another:

B

A D C

then the straight line DA is equal to DB, (1. 6.) and therefore, to DC; and because the three straight lines DA, DB, DC are all equal, therefore D is the center of the circle. (III. 9.)

From the center D, at the distance of any of the three DA, DB, DC, describe a circle;

this shall pass through the other points;

and the circle of which ABC is a segment has been described: