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G CE

Find E, the centre of the circle, and join AE; and from the centre E, at the distance EA, describe the circle AFG; from the point D draw DF at right angles to EA, and join EBF, AB: AB touches the circle BCD. For, because E is the centre of the circles AFG, BCD, EA is equal to EF, and ED to EB: Therefore the two sides AE, EB are equal to the two FE, ED, and the angle at E is common to the two triangles AEB, FED-therefore the base AB is equal to the base FD, and the triangle AEB to the triangle FED, and the other angles to the other angles, each to each, to which the equal sides are opposite; therefore the angle ABE is equal to the angle FDE: But FDE is a right angle, therefore also ABE is a right angle: And EB is drawn from the centre; but (3. 16. Cor.) a straight line, drawn at right angles to the diameter of a circle, from the extremity of it, touches the circle; therefore AB touches the circle, and it is drawn from the given point A.

But if the given point be in the circumference of the circle, as D, draw DE to the centre E, and DF at right angles to DE: DF touches the circle.

PROP. XVIII. THEOR.

Q. E. F.

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: FC is perpendicular to DE.

For, if not, from the point F draw FBG perpendicular to DE: Then, because FGC

D

A

1

is a right angle, therefore FCG must be less than a right angle (1. 17): and to the greater angle the greater side is opposite; therefore FC is greater than FG: But FC is equal to FB; therefore also FB is greater than FG, the less than the greater-which is absurd: Therefore FG is not perpendicular to DE: And, in like manner, it may be shewn that no other line than FC can be perpendicular to DE, that is, FC is perpendicular to DE.

Wherefore, If a straight line &c.

Q. E. D.

PROP. XIX. THEOR.

If a straight line touch a circle, and from the point of contaet 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: the centre of the circle shall be in CA.

For, if not, let, if possible, F be the centre, and join CF: Then, because DE touches the

B

C

E

circle ABC, and FC is drawn from the centre to the point of contact, FC is perpendicular to DE (3. 18), and therefore FCE is a right angle: But ACE is also a right angle; therefore the angle FCE D is equal to the angle ACE, the less to the greaterwhich is absurd: Therefore F is not the centre of the circle ABC: And, in like manner, it may be shewn that no other point, which is not in CA, can be the centre, that is, the centre is in CA.

Wherefore, If a straight line &c. Q.E.D.

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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 the same part of the circumference, BC, for their base: the angle BEC shall be double of the angle BAC.

E

First, let E, the centre of the circle, be within the angle BAC; join AE, and produce it to F: Then, because EA is equal to EB, the angle EAB is equal to the angle EBA; and therefore the angles EAB, EBA are double of the angle EAB: But the angle BEF (1. 32) is equal to the angles EAB, EBA; therefore also the angle BEF is double of the angle EAB: In like manner, the angle FEC is double of the angle EAC: Therefore the whole angle BEC is double of the whole angle BAC.

B

F

C

Next, let E, the centre of the circle, be without the angle BAC: join AE, and produce it to F: Then, as in the first case, it may be shewn that the

angle FEC is double of the angle FAC, and that the angle FEB, a part of the first, is double of the angle FAB, a part of the other; therefore the remaining angle BEC is double of the remaining angle ВАС.

F

Wherefore, The angle at the centre &c. Q.E.D.

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B

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: the angles BAD, BED shall be equal to each other.

A B

Take F the centre of the circle ABCD: and, first, let the segment BAED be greater than a semicircle, and join BF, FD: Then, because the angle BFD at the centre and the angle BAD at the circumference have the same part of the circumference, viz. BCD, for their base; therefore (3. 20)

the angle BFD is double of the angle BAD: In like manner, the angle BFD is double of the angle BED: Therefore the angle BAD is equal to the angle BED. Next, let the segment BAED be not greater than a semicircle: draw AF to the centre,

and produce it to C, and join CE: Then B
the segment BAEC is greater than a
semicircle; and the angles in it, BAC,
BEC, are equal, by the first case: For
the same reason, because the segment

E

D

CAED is greater than a semicircle, the angles CAD, CED are equal: Therefore the whole angle BAD is equal to the whole angle BED.

Wherefore, The angles in the same segment &c. Q. E. D.

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

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

Join AC, BD: Then, because the three angles of every triangle are together equal to two right angles,

the three angles of the triangle BAC, viz. ACB, BAC,

A

D

C

B

CBA, are together equal to two right angles: But the angle ADB is equal to the angle ACB (3. 21), because they are in the same segment ADCB; and the angle BDC is equal to the angle BAC, because they are in the same segment BADC; therefore the whole angle ADC is equal to the angles ACB, BAC: To each of these equals add the angle ABC; therefore the angles ADC, ABC are equal to the angles ACB, BAC, CBA: But ACB, BAC, CBA are together equal to two right angles; therefore also the angles ABC, ADC are together equal to two right angles: And, in like manner, it may be shewn that the angles BAD, BCD, are together equal to two right angles. Wherefore, The opposite angles &c. Q. E.D.

PROP. XXIII. THEOR.

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.

D

C

B

If it be possible, let two similar segments of circles, viz. ACB, ADB, be upon the same side of the same straight line AB, not coinciding with one another: Now, because the circle ACB cuts the circle ADB in the A two points A, B, they cannot cut one another in any other point (3. 10); and therefore one of the segments must fall within the other; let ACB fall within ADB, and draw the straight line BCD, and join AC, AD.

Then, because the segment ACB is similar to the segment ADB, and that similar segments of circles contain equal angles, the angle ACB is equal to the angle

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