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

2. The base DF is equal to the base AB (I. 4), and the triangle FED to the triangle AEB;

and the other angles to the other angles; therefore

3. The angle EBA is equal to the angle EDF;

but EDF is a right angle (constr.), therefore

4. EBA is a right angle (Ax. 1),

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

5. 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 centre E, and DF at right angles to DE; then

1. DF touches the circle (III. 16. Cor.)

Q.E.F.

PROPOSITION 18.-Theorem.

If a straight line touch a circle, the straight line drawn from the centre 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 centre F, and draw the straight line FC (III. 1).

Then FC shall be perpendicular to DE.

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

Demonstration. And because FGC is a right angle, therefore

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

E

and to the greater angle the greater side is opposite (I. 19); therefore 2. FC is greater than FG;

but FC is equal to FB (I. Def. 15); therefore

3. FB is greater than FG;

the less than the greater, which is impossible; therefore

4. FG is not perpendicular to DE.

In the same manner it may be shown, that no other line is perpendicular to DE beside FC; that is,

5. FC is perpendicular to DE.

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

PROPOSITION 19.-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 centre of the circle shall be in that line.

Let the straight line DE touch the circle ABC in C, and from C let CA be drawn at right angles to DE.

Construction. For, if not, let F be the centre, if possible, and join CF.

Demonstration. Because DE touches the circle ABC, and FC is drawn from the centre to the point of contact, therefore

therefore

1. FC is perpendicular to DE (III. 18);

2. FCE is a right angle;

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

3. The angle FCE is equal to the angle ACE ((Ax. 1);

the less to the greater, which is impossible; therefore

4. F is not the centre of the circle ABC.

In the same manner it may be shown, that no other point which is not in CA, is the centre; that is,

5. The centre of the circle is in CA.

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

PROPOSITION 20.-Theorem.

The angle at the centre 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 centre, 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.

First. Let the centre of the circle be within the angle BAC.
Construction. Join AE, and produce it to F.

Demonstration. Because EA is equal to EB, therefore the angle EBA is equal to the angle EAB (I. 5); therefore

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

2. The angle BEF is double of the angle EAB;

for the same reason,

therefore

3. The angle FEC is double of the angle EAC;

4. The whole angle BEC is double of the whole angle BAC. Secondly. Let the centre of the circle be without the angle BAC. Construction. Join AE, and produce it to meet the circumference

in F.

Demonstration. It may be demonstrated, as in the first case, that 1. The angle FEC is double of the angle FAC,

and that

therefore

2. FEB a part of FEC; is double of FAB, a part of FAC;

3. The remaining angle BEC is double of the remaining angle BAC.

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

H

PROPOSITION 21.-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.

Construction. Take F, the centre of the circle ABCD (III. 1), and join BF, FD.

Demonstration. Because the angle BFD is at the centre, 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 1. The angle BFD is double of the angle BAD (III. 20) ;

for the same reason,

therefore

2. The angle BFD is double of the angle BED;

3. The angle BAD is equal to the angle BED (Ax. 7). Next. Let the segment BAED be not greater than a semicircle.

Construction. Draw AF to the centre, and produce it to C, and join CE.