Because F is the centre of the circle ABC (I. Def. 15), FA is equal to FC. Because G is the centre of the circle ADE, GA is equal to GD. Therefore FA and AG are together equal (I. Ax. 2) to FC and DG; and the whole FG is greater (I. Ax. 9) than FA and AG. But FAG is a triangle, and FG is also less than FA and AG (I. 20);

which is impossible. Therefore the straight line which joins the centres of the circles BAC and EAD cannot pass otherwise than through the point A; that is, it must pass through the point A. Therefore, if two circles, &c. Q. E. D.

H

PROP. XIII. (THEOREM.)-One circle (EBF) cannot touch another (ABC) in more points than one (B), either internally or externally. For, if it be possible, let EBF touch ABC in another point D. Join BD, and draw GH bisecting (I. 11) BD at right angles. Because the points B and D are in the circumference of both circles, the straight line BD lies (III. 2) within each of them. Therefore their centres are (III. Cor. 1) in the straight line GH which bisects BD at right angles. Because GH joins their centres it passes through (III. 11) the points of contact B

E

H B

D

and D. Therefore G H coincides with DB. But GH is also at right angles to BD (Const.); which is impossible. Therefore one circle EBF cannot touch another ABC, internally, in more points than one. Again, let the circle ACK touch the circle ABC externally in the point A. ACK cannot touch ABC in any other point.

For, if it be possible, let ACK touch ABC in another point C. Join AC.

Because the two points A and C are in the circumference of the circle ACK, the straight line AC which joins them, lies within (III. 2) the circle ACK. But the circle ACK is without (Hyp.) the circle ABC. Therefore the straight line AC is without the circle ABC. Because the points A and C are in the circumference of the circle ABC, the straight line AC lies (III. 2) within the circle ABC. But it has been proved that AC also lies without the circle

A

K

ABC; which is absurd. Therefore one circle ACK cannot touch another circle ABC, externally, in more points than one. Therefore, one circle, &c. Q. E. D.

PROP. XIV. (THEOREM.)-Equal straight lines (AB, CD) in a circle (BACD) are equally distant from the centre; and, conversely, straight lines equally distant from the centre are equal to one another.

Let the straight lines AB and CD, in the circle BACD, be equal to

one another. They are equally distant from the centre.
Take E, the centre of the circle A BDC (III. 1),
and from E draw EF and EG perpendiculars to AB
and CD (I. 12) respectively. Join EA and EC.

A

F

B

E

Because the straight line EF, passing through the centre, cuts the straight line AB, which does not pass through the centre, at right angles (III. 3), AB is bisected at F. Therefore AF is equal to FB, and AB double of AF. For the same reason, CD is double of CG. But AB is equal (Hyp.) to CD. Therefore AF is equal (I. Ax.7) to CG, and their squares are equal. Because AE is equal to EĆ (I. Def. 15), the square of AE is equal to the square of EC. But the squares of AF and FE are equal (I. 47) to the square of AE. Also, the squares of EG and GC are equal (I. 47) to the square of EC. Therefore the squares of AF and FE are equal (I. Ax. 1) to the squares of CG and GE. From these equals take the squares of AF and CG, which were shown to be equal. Therefore the remaining square of EF is equal (I. Ax. 3) to the remaining square of EG, and the straight line EF to the straight line EG. But straight lines in a circle are said to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal (III. Def. 4). Therefore AB and CD are equally distant from the centre.

Next, let the straight lines AB and CD be equally distant from the centre (III. Def. 4); that is, let FE be equal to EG. The straight lines AB and CD are equal.

For, the same construction being made, it may, as before, be demonstrated that AB is double of AF, end CD double of CG, that the squares of FE and EG are equal, and that the squares of EF and FA are equal to the squares of EG and GC. From these equals take the squares of FE and EG, which are equal. Therefore the remaining square of AF is equal (I. Ax. 3) to the remaining square of CG; and the straight line AF to the straight line CG. But AB was shown to be double of AF, and CD double of CG. Therefore AB is equal (I. Ax. 6) to CD. Therefore equal straight lines, &c. Q. E. D.

PROP. XV. (THEOREM.)-The diameter (AD) is the greatest straight line in a circle (ABCD); and, of all others, that (BC) which is nearer to the centre (E) is greater than (FG) one more remote: and the greater is nearer to the centre than the less.

From the centre E draw EH and EK perpendiculars to BC and FG (I. 12), respectively, and join EB, EC, and EF.

F

K

A B

H

E

Because AE is equal (I. Def. 15) to EB, and ED to EC; therefore AD is equal (I. Ax. 2) to EB and EC; but EB and EC are greater (I. 20) than BC. Therefore also AD is greater than BC. Because BC is nearer (Hyp.) to the centre than FG, EH is less (III. Def. 5) than EK. Therefore the square of EH is less than the square of EK. But it may be shown, as in the preceding proposition, that BC is double of BH, and FG double of FK, and that the squares of EH and HB are

D

equal to the squares of EK and KF. But the square of EH is less than the square of EK. Therefore the square of BH is greater than the square of FK, and the straight line BH greater than the straight line FK. Therefore, also, BC is greater than FG.

Next, let BC be greater than FG. The greater, BC, is nearer to the centre than the less FG; that is, EH is less than EK.

For, let the same construction be made. Because BC is greater than FG (III. Def. 5), BII is greater than KF, and the square of BH is greater than the square of FK. But, as before, the squares of BH and HE are equal to the squares of FK and KE. Therefore the square of EH is less than the square of EK, and the straight line EH less than EK. Therefore BC is nearer (III. Def. 5) to the centre FG. Wherefore the diameter, &c. Q. E. D.

PROP. XVI. (THEOREM.)-If a straight line (AE) pass through one of the extremities of the diameter (AB) of a circle (ABC), at right angles to the diameter, it touches the circumference; but if it pass through the same point not at right angles to the diameter, it cuts the circumference.

In AE take any point F, and join DF. Let DF meet the circumference at C.

B

E

F

Because, in the triangle DAF, the angle DAF (Hyp.) is a right angle; therefore the angle DAF is greater than (1.17) the angle DFA, and DF is greater than DA (I. 19). But DA is equal to DC (I. Def. 15). Therefore DF is greater than DC, and the point F (III. Ax. 1) is without the circumference. In the same manner it may be shown that any point in AE, but the point A, is without the circumference. Therefore A E lies wholly without the circumference, except at the point A. Wherefore AE touches the circumference at the point A.

Next, let the straight line AC pass through the point A, the extremity of the diameter, not at right angles to AB. The straight line AC cuts the circumference of the circle ABC.

From the centre D draw DF (I. 12) at right angles to AC, and let it meet the circumference at G.

E

G

A

Because, in the triangle DAF, the angle D FA is B a right angle (Const.), the angle DAF (I. 17) is less thau the angle DFA. Therefore DF is less than DA (I. 19). But DA is equal to DG (I. Def. 15). Therefore DF is less than DG, and the point F is within the circumference (III. Ax. 1). But the point F is in the straight line AC. Therefore the straight line AC cuts the circumference. Therefore, if a straight line pass, &c. Q. E. D.

PROP. XVII. (PROBLEM.)-To draw a straight line from a given point, either in the circumference (D) of a given circle (BDC) or without it (A), that shall touch the circumference.

First, when the point A is without the cirumference. Find the centre of the circle BCD, and join AE. From the centre E, at the

distance EA, describe the circle AFG. From the point D, draw DF at right angles to EA (I. 11). Join EF and AB. The straight line AB touches the circle BCD at the point B.

(C E

Because E is the centre of the circles BCD and G AFG, EA is equal to EF (I. Def. 15) and ED to EB; therefore, in the two triangles AEB, FED, the two sides AE and EB are equal to the two sides FE and ED, each to each; and they contain the angle at E common to the two triangles AEB and FED. Therefore the base DF is equal to the base AB (I. 4) the triangle FED to the triangle AEB, and the remaining angles to the remaining angles of the other, each to each. Therefore the angle EBA is equal to the angle EDF. But EDF is a right angle (Const.) Therefore EBA is a right angle (I. Ax. 1). But a straight line which passes through the extremity of a diameter (or radius), at right angles to it, touches the circle (III. 16). Therefore AB touches the circle; and it is drawn from the given point A.

Next, when the given point D is in the circumference of the circle. Find the centre E as before. Join DE, and draw DF at right angles to DE (I. 11). For the same reason as above DF touches the circle (III. 16), and it is drawn from the given point D. Q. E. F.

PROP. XVIII. (THEOREM.-If a straight line (DE) touch a circle (ABC), the straight line drawn from the centre (F) to the point of contact (C) is perpendicular to the line touching the circle.

For, if FC be not perpendicular to DE; from the point F draw FG perpendicular to DE, and let it meet the circumference in B.

Because FGC is a right angle, GCF is an acute angle (I. 17). But in any triangle, the greater side is opposite (I. 19) to the greater angle. Therefore FC is greater than FG. But FC is equal to FB (1. Def. 15). Therefore

D

A

C

B

GE

FB is greater than FG; that is, a part greater than the whole, which is impossible. Therefore FG is not perpendicular to DE. In the same manner it may be shown, that no other straight line can be perpendicular to DE but FC. Therefore FC is perpendicular to DE. Therefore, if a straight line, &c. Q. E. D.

PROP. XIX. (THEOREM.)-If a straight line (DE) touch a circle (ABC), and from the point of contact (C) a straight line (AC) be drawn at right angles to the tangent, the centre of the circle shall be in that straight line (CA).

For, if the centre of the circle be not in CA, let, if possible, F, a a point out of CA, be the centre, and join CF.

Because DE touches the circle ABC at the point C, and FC is drawn from the centre to the point of contact; therefore FC is perpendicular to DE (III. 18), and FCE is a right angle. But

ACE is also (Hyp.) a right angle. Therefore the angle FCE is equal (I. Ax. 1) to the angle ACE; that is, the less to the greater, which is impossible. Therefore F is not the centre of the circle ABC. In the same manner it may be shown that no other point out of CA is the centre. Therefore the centre of the circle is in CA. Therefore, if a straight line, &c. Q. E. D.

PROP. XX. (THEOREM.)-The angle (BEC) at the centre (E) of a circle (ABC) is double of the angle (BAC) at the circumference standing upon the same arc, that is, upon the same part of the circumference (BC). Join AE, and produce it to F. First, let the centre of the circle be within the angle BAC.

Because EA is equal to EB, the angle EAB is equal (I. 5) to the angle EBA. Therefore the two angles EAB and EBA are double of the angle EAB. But the angle BEF is equal (1.32) to the two angles EAB and EBA. 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 (III. Ax. 2) of the whole angle BAC.

Next, let the centre of the circle be without the angle BAC.

It may be demonstrated, as in the preceding 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 (III. Ax. 3) of the remaining angle BAC. Therefore the angle at the centre, &c. Q. E. D.

B

F

E

PROP. XXI. (THEOREM.)-The angles (BAD and BED) in the same segment of a circle (ABCD) are equal to one another. Take F, the centre of the circle ABCD (III. 1), and join BF and FD..

Because the angle BFD at the centre, and the angle BAD at the circumference, stand upon the same arc BCD, the angle BFD is double (III. 20) of the angle BAD. For the same reason the angle BFD is double of the angle BED. Therefore the angle BAD is equal (I. Ax. 7) to the angle BED. B Next, let the segment BAED be not greater than a semicircle.

Join AF, produce it to C, and join CE. Because the segment BADC is greater than a semicircle, the angles BAC and BEC are equal, by the first case. Because CBED is greater than a semicircle, the angles CAD and CED are also equal, by the first case. Therefore the whole angle BAD is equal (I. Ax. 2) to the whole angle BED. Wherefore the angles in the same segment, &c. Q. E. D.

E

B

A

A

F

E

D