The first three books of Euclid's Elements of geometry, with theorems and problems, by T. Tate

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 (1. 12.) DG perpendicular to FA, and let it meet the circumference in H: And because AGD is a right angle, and DAG less (1. 17.) than a right angle: DA is greater (1. 19.), than DG: But DA is equal to DH; therefore DH is greater than DG, the less than the greater, which is impossible: Therefore 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 A, 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; and that it touches it only in one point, because, if it did meet the circle in two, it would fall within it (II. 2.). "Also it is evident that there can be but one straight line which touches the circle in the same point."

PROP. XVII. PROB.

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

First, let a be a given point without the given circle BCD, it is required to draw a straight line from a 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 (1. 11.) FD at right angles to AE; and join EBF, AB.

AB touches the circle BCD.

Because E is the centre of the circles BCD, AFG, EA is

equal to EF, and ED to EB; therefore the two sides A E, EB are equal to the two FE, ED, and they contain the angle at E common to the two triangles AEB, FED; therefore 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 (1. 4.) : Therefore the angle EBA is equal

to the angle EDF: But EDF is a right angle, wherefore 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 (m. 16. Cor.): Therefore AB touches the circle; and it is drawn from the given point A. 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; DF touches the circle (m. 16. Cor.).

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 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, GCF is (1. 17.) an acute angle; and to the greater angle the greatest (1. 19.) side is opposite; therefore FC is greater than FG; but FC is equal to FB; therefore FB is greater than FG, the less than the greater which is impossible: Wherefore FG is not perpendicular to DE: In the same manner it may be shown, that no other is perpendicular to it besides FC, that is, FC is perpendicular to DE. Therefore, if a straight line, &c. Q. E. D.

If a straight line touches 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; 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, FC is perpendicular (m. 18.) to B DE; therefore FCE is a right angle: But ACE is also a right angle; therefore the angle FCE is equal to the angle ACE, the less to the greater, which is impossible: Wherefore F is not the centre of the circle

ABC. In the same manner, it may be shown, D C that no other point which is not in CA, is the centre; that is, the centre is in CA. Therefore if a straight line, &c. Q. E. D.

PROP. XX. THEOR.

The angle at the centre of the 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.

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

EBA; therefore also the angle BEF is double of the angle FAB: For the same reason, the angle FEC is double of the angle EAC: Therefore the whole angle BEC is double of the whole angle BAC.

Again, let E the centre of the circle be without the angle BDC, and join DE and produce it to G. It may be demonstrated, as in the first case, that the angle GEC is double of the angle GDC, and that GEB a part of the first is double of GDB a part of the other; therefore the remaining angle BEC is double of the remaining angle BDC. Therefore the angle at the centre, &c. Q. E. D.

PROP. XXI. THEOR.

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 are equal to one another.

Take F the centre of the circle ABCD: And, first, let the segment BAED be greater than a semicircle, and join BF, FD: And 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. BCD, for their base; therefore 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 to the angle BED.

A E

But if the segment BAED be not greater than a semicircle, let BAD, BED be angles in it; these also are equal to one another: Draw AF to the centre, and produce it to c, and join CE: B Therefore the segment BADC is greater than a semicircle; and the angles in it BAC, BEC are equal, by the first case: For the same reason, because CBED 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.

PROP. XXII. THEOR.

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

Let ABCD be a quadrilateral figure in the circle ABCD; any two of its opposite angles are together equal to two right angles.

Join AC, BD; and because the three angles of every triangle are equal (1. 32.) to two right angles, the three angles of the triangle CAB, viz. the angles CA B, ABC, BCA, are equal to two right angles: But the angle CAB is equal (III. 21.) to the angle CDB, because they are in the same segment BADC, and the angle ACB is equal to the angle ADB, because they are in the same segment ADCB: Therefore the whole angle ADC is equal to the angles CAB, ACB: To each of these equals add the angle ABC; therefore the angles ABC, CAB, BCA are equal to the angles ABC, ADC: But ABC, CAB, BCA are equal to two right angles; therefore also the angles ABC, ADC are equal to two right angles: In the same manner, the angles BAD, DCB may be shown to be equal to two right angles. Therefore, 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.

If it be possible, let the two similar segments of circles, viz. ACB, ADB, be upon the same side of the same straight line AB, not coinciding with one another: Then, because the circle ACB cuts the circle ADB in the two points A, B, they cannot cut one another in any other point (11. 10.): One of the segments must therefore fall within the other: let ABC fall within ADB, and draw the straight line BCD, and join CA, DA: And because the segment ACB is similar to the segment ADB, and that similar segments of circles contain (ш. Def. 11.) equal angles; the angle ACB is equal to the angle ADB, the

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The first three books of Euclid's Elements of geometry, with theorems and ...