but, as was demonstrated in the preceding proposition,

2.

and

BC is double of BH, and FG double of FK,

3.

The squares of EH, BH, are equal to the squares of

EK, KF,

of which the square of EH is less than the square of EK, because EH is less than EK; therefore

4. The square of BH is greater than the square of FK, and the straight line BH greater than FK;

and therefore

5. BC is greater than FG.

Next, let BC be greater than FG; BC is nearer to the centre than FG, that is, the same construction being made, EH is less (III. Def. 5.) than EK. Because BC is greater than FG, likewise

2.

The squares of BH, HE, are equal to the squares of

FK, KE,

of which the square of BH is greater than the square of FK, because BH is greater than FK; therefore

3. The square of EH is less than the square of EK, and the straight line E less than EK.

Wherefore, the diameter, &c.

Q.E.D.

PROP. XVI.-THEOREM.

The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle; and no straight line can be drawn between that straight line and the circumference from the extremity, so as not to cut the circle; or, which is the same thing, no straight line can make so great an acute angle with the diameter at its extremity, or so small an angle with the straight line which is at right angles to it, as not to cut the circle.

Let ABC be a circle, the centre of which is D, and the diameter AB: the straight line drawn at right angles to AB from its extremity A, shall fall without the circle.

For, if it does not, let it fall, if possible, within the circle, as AC; and draw DC to the point C where it meets the circumference. And because DA is equal (I. Def. 15.) to DC, (I. 5.)

1. The angle DAC is equal the angle ACD;

but DAC is (Hyp.) a right angle, therefore ACD is a right angle, and therefore

2. The angles DAC, ACD, are equal to two right angles; which is impossible (I. 17.): therefore

3. The straight line drawn from ▲ at right angles to BA does not fall within the circle.

In the same manner it may be demonstrated that it does not fall upon the circumference; therefore

4. The straight line drawn from A at right angles to BA

must fall without the circle, as AE.

And between the straight line AE and the circumference no straight line can be drawn from the point A which does not cut the circle. For, if possible, let FA be between them, and from the point D draw (I. 12.) DG perpendicular to FA, and let it meet the circumference in H.

F

G

H

B

And because AGD is a right angle, and DAG less (I. 17.) than a right angle, (I. 19.)

1. DA is greater than DG:

but DA is equal to DH; therefore

2. DH is greater than DG,

the less than the greater, which is impossible: therefore

3. No straight line can be drawn from the point A, between AE and the circumference, which does not cut the circle:

or, which amounts to the same thing, however great an acute angle a straight line makes with the diameter at the point 4, or however small an angle it makes with AE, the circumference passes between the straight line and the perpendicular AE.

"And this is all that is to be understood, when, in the Greek text and translations from it, the angle of the semicircle is said to be greater than any acute rectilineal angle, and the remaining angle less than any rectilineal angle.”

COR. From this it is manifest that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it, touches the

circle (III. Def. 2.); and that it touches it only in one point, because, if it did meet the circle in two, it would fall within it (III. 2.). "Also it is evident that there can be but one straight line which touches the circle in the same point."

PROP. XVII.-PROBLEM.

To draw a straight line from a given point, either without or in the circumference, which shall touch a given circle.

First, let 4 be a given point without the given circle BCD; it is required to draw a straight line from 4 which shall touch the circle.

Find (III. 1.) the centre E of the circle, and join AE; and from the centre E, at the distance AE, describe the circle AFG; from the point D draw (I. 11.) FD at right angles to AE; and join EBF, AB: AB touches the circle BCD.

A

Because E is the centre of the circles BCD, AFG; (I. Def. 15.)

1. 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 (I. 4.)

2. The base DF is equal to the base AB, 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 (Constr.) a right angle, wherefore

4. EBA is a right angle:

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 4. Which was to be done.

But 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; (III. 16. Cor.)

1. DF touches the circle.

PROP. XVIII.- -THEOREM.

If a straight line touches 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 (III. 1.) the centre F, and draw the straight line FC; FC is perpendicular to DE.

For, if it be not, from the point F draw FBG perpendicular to DE; and because FGC is a right angle, (I 17.)

1. GCF is an acute angle;

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

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

the less than the greater, which is impossible; wherefore

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

5. FC is perpendicular to DE.

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

PROP. 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 centre 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; the centre of the circle is in CA.

For, if not, let F be the centre, if possible, and join CF. Because DE touches the circle ABC, and FC is drawn from the centre to the point of contact, (III. 18.)

therefore

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,

the less to the greater, which is impossible. Wherefore 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.

PROP. XX.-THEOREM.

The angie 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 the same circumference BC for their base; the angle BEC is double of the angle BAC.

B
F

First let E the centre of the circle be within the angle BAC, and join AE, and produce it to F. Because EA is equal to EB, the angle EAB is equal (I. 5.) to the angle EBA; therefore

1. The angles EAB, EBA, are double of the angle EAB; but the angle BEF is equal (I. 32.) to the angles EAB, EBA; 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. Again, let E the centre of the circle be without the angle BAC.

B