PROP. XVII. PROB.

To draw a tangent to a given circle (BGE) from a given point, either in or outside of its circumference.

A

If the given point be in the circumference, as G, it is plain that the straight line drawn from G perpendicular to the ray CG, will be the required tangent (iii. Prop. 16).

F

But if the given point be outside of the circumference, as A, join CA, cutting the E circumference of the given circle in B; with

B

the centre C and ray CA describe the circle ADF; from B draw BD perpendicular to CA; join CD, cutting the circumference of the given circle in G, and draw AG, which will be the required tangent.

For the triangles CGA, CBD have the sides CA = CD and CG=CB, and the angle contained by those equal sides is common to both triangles; the remaining angles therefore opposite to equal sides, are also equal (i. Prop. 4), and CGA =/ CBD; but CBD is a right angle (Const.), and therefore CGA is also a right angle, and consequently, AG drawn from the given point and perpendicular to the diameter of the given circle at its extremity, is a tangent to the circle (iii. Prop. 16).

If a straight line (FE) touches a circle, the ray (CD) drawn to the point of contact (D) is perpendicular to the tangent.

A

B

For, if it be supposed that CD is not perpendicular to FE, but that some other straight line, as CF, drawn from the centre is perpendicular to it; then, since CFD is supposed to be a right angle, CDF must be less than a right angle (i. Prop. 32, Cor.), and consequently CF must be less than CD (i. Prop. 19): but since the

F

straight line FE touches the circle at D, the point F must lie outside of the circumference (iii. Prop. 16), and therefore CF must be also greater than CD, which is absurd. Therefore

CF is not perpendicular to FE; and the like may be proved of every line drawn from the centre of the circle except CD, which is drawn to the point of contact.

If a straight line (ED) touches a circle, the straight line (BA) drawn from the point of contact (B) perpendicular to the tangent, passes through the centre of the circle. For, if it be supposed that the centre of the circle is not in the straight line BA drawn perpendicular to the tangent from the point of contact, but that it is a point outside of that line, as C; join CB, and then, because CB is supposed to be a ray drawn to the point of contact, it must be perpendicular to the tangent (iii. Prop. 18), and CBD must be a right angle; but ABD is also a right angle (Hyp.), therefore

CBD=/ ABD, a part equal to the whole, which is absurd. Therefore C is not the centre of the circle; and the same may be proved of any other point outside of BA, which therefore passes through the centre.

The angle (ACB) at the centre of a circle, is double of the angle (ADB) at the circumference, standing on the same arch (AB).

D

First, suppose that C, the centre of the circle, is in AD, one of the legs of the angle at the circumference. In this case it is manifest that Z ACB is equal to s CDB, CBD together (i. Prop. 32); but these angles are equal to each other, since CB = CD (i. Prop. 5); therefore ZACB, or the angle at the centre, is double of CDB, that is to say, of ▲ ADB, or the angle at the circumference.

B

Secondly, let C, the centre of the circle, be within the angle ADB; join DC, and produce it to E, and also draw CG parallel to DB, and CF parallel to DA. Then, because

B

CD=CB, ▲ CBD= / CDB (i. Prop. 5); but ▲GCB=/ CBD and / ECG = / CDB (i. Prop. 29), and therefore / GCB=/ ECG, and ▲ ECB is bisected by CG and in like manner it may be shown that ACE is bisected by CF; therefore Zs ACE, ECB are, taken together, double of /s FCE, ECG; but /s FCE, ECG are respectively equal to s CDA, CDB (i. Prop. 29); and are therefore, taken together, equal to ▲ ADB, and therefore the Zs ACE, ECB taken together, that is to say, the angle ACB at the centre, being double of s FCE, ECG, is also double of the angle ADB at the circumference.

G

D

Thirdly, let C, the centre of the circle, be outside of the angle ADB ; join DC, as before, and produce it to E; draw CG parallel to DB, and CF parallel to DA. In this, as in the preceding case, the angle ECB is bisected by CG, and the angle ECA by CF; therefore ECB

= 2 ▲ ECG, and ▲ ECA = 2 ≤ ECF, and,

=

=

E

F

B

A

taking the latter from the former, ECB ▲ ECA 2 / ECG 2 / ECF; or the difference between ECB and ECA, that is to say, ACB, is double of the difference between / ECG and ECF, or / FCG: and since s ECG, ECF are respectively equal to s EDB, EDA, their differences are also equal (Ax. 3), or ≤ FCG = ADB; but F

it has been proved that ACB, or the angle

at the centre, is double of ▲ FCG, it is there

C

fore also double of ADB, or the angle at the circumference standing on the same arch.

All angles (BAD, BED) in the same segment (BAED) of a circle are equal to one another.

First, let the segment BAED be greater than a semicircle: from C the centre of the circle draw CB, CD; then / BCD, because it is at the centre, is double of ▲ BAD at the circumference (iii. Prop. 20); and in like manner it is double of BED, and therefore BAD =BED (Ax. 7).

B

B

E

D

But if the segment BAED be not greater than a semicircle, draw AC to the centre, and produce it to the circumference in F, and join EF; and then, because ADF is a semicircle, the segment BADF is greater than a semicircle; and, as has been above demonstrated, the angles contained in it are equal, and

BAF

F

=/ BEF; and in like manner / FAD= / FED; therefore the whole BAD=/ BED (Ax. 2).

PROP. XXII. THEOR.

The opposite angles (A and D, B and c) of any quadrilateral figure (ACDB) inscribed in a circle, are together equal to two right angles.

Draw the diagonals AD, BC; and then because angles in the same segment are equal, ▲ ACB=/ ADB,

+▲ BAD; add to both ABD, and then

▲ ACD + 2 ABD, or two opposite angles of a the quadrilateral figure are equal to ADB

B

+2 BAD +2 ABD, or the three angles of a triangle, that is to say, they are equal to two right angles. In like manner it may be shown that s BAC and BDC are together equal to two right angles.

side

Upon the same straight line (AB), and upon the same of it, there cannot be two similar segments of circles not coinciding with each other.

For, if two segments of circles be upon the line AB, and

D

C

at the same side of it, since they meet at the points A and B, they cannot meet in any other point (iii. Prop. 10), and therefore one of them must fall within the other. Draw BD to the circumference of the outer segment, cutting the circumference of the inner segment in C, and join AC and AD. Then, if the segments be supposed similar,

A

ZACB=/ ADB (iii. Def. 12); but ACB is also greater than ADB (i. Prop. 16), which is absurd. Therefore the segments not coinciding, are not similar, and, consequently, there cannot be on the same straight line, and on the same side of it, similar segments of circles, which do not coincide.

Similar segments (AEB, CFD) of circles upon equal straight lines (AB, CD) are equal to each other.

E

F

For, if the segment AEB be so applied to the segment CFD, that the point A may be on C, the line AB may fall on CD, and the arches AEB, CFD may lie at the same side; then because AB-CD, the point B must fall on D, and the line AB must wholly coincide with CD; and the similar segments AEB, CFD, being then upon the same straight line and at the same side of it, must also coincide (iii. Prop. 23), and are therefore equal.

C

D

PROP. XXV. PROB.

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

Bisect AB in D, and from the point D draw to the circumference of the segment, DE perpendicular to AB, and join AE. Then if EAD = / AED, it is plain that

ED = AD (i. Prop. 6) = DB (Const.), and therefore the point D from which more than two equal straight lines are drawn to the circumference must be the centre of the circle (iii. Prop. 9). If

from D, therefore, as centre, and with either of those lines ED, AD, or BD for interval, a circle be described, it will pass through the points A, E and B, and will be the circle of which AEB is a segment.

But if s EAD, AED be not equal, then at the point A, with the line EA make the angle EAC equal to /AED (i. Prop. 23), and produce, if necessary, ED till it meets AC, and join