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 the square of BH is greater than the square of FK, and the straight line BH greater than FK; and therefore 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 than EK: because BC is greater then FG, BH is likewise greater than KF : and 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 the square of EH is less than the square of EK, and the straight line EH less than EK. Wherefore the diameter, &c. Q. E. D.

PROP. VIII. THEOR.

A perpendicular drawn from the extremity of a radius, is a tangent to the circumference of a circle.

Let DFG be a circle, A the centre, AD a radius, and BC perpendicular to AD at the point D; BC is a tangent to the circumference DFG. Take E any point in BC, and join AE.

Because AD is perpendicular to BC, AE is oblique to BC, and therefore AE is longer than AD (13. 1.); and therefore the point E is without the circumference. It may be shewn in the same way, that

any point other than D, is without the circumference; and therefore BC does not cut the circumference; and therefore it is a tangent (Def. 1. 3.). Q. E. D.

COR. Any other straight line through D, as HI, will cut the cir

cumference.

Because AD is perpendicular to BC, and HI does not coincide with BC, AD is oblique to HI. Draw AH perpendicular to HI (10. 1); and because AH is perpendicular, and AD oblique to HI, AD is longer

than AH (12, 1.); and therefore the point H falls within the circumference; and therefore the straight line HI cuts the circumfe

rence.

It may be shewn in the same way, that any other straight line through D, except BC, will cut the circumference. Q. E. D.

PROP. IX. THEOR.

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 (3. 3.) 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 (9. 1.) DF at right angles to EA, 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 AE, EB are equal to the two FE, ED, and G 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 (4. 1.); therefore the angle EBA is equal to the angle EDF; but EDF is a right angle, wherefore EAB is a right angle, and EB is drawn from

C

E

B

the centre; but a straight line drawn from the extremity of a diameter, at right angles to it, touches the circle (8. 3.); 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.

PROP. X. THEOR.

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 it be not, from the point F draw FBG perpendicular to DE (13. 1.); then the oblique line FC is longer than the perpendicular FG (13. 1.); 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 D perpendicular to DE. Therefore, if a straight line, &c. Q. E. D.

PROP. XI. THEOR.

A

F

B

GE

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 C let CA be drawn at right angles to DE; the centre of the circle is in CA.

B

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 (10. 3.) to 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, 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.

D

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 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 (5. 1.) to the angle EBA; therefore the angles EAB, EBA are double of the angle EAB; but the angle BEF is equal (5. 2.) to the angles EAB, 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 of the whole angle BAC.

Again, Let BDC be inflectedt o the circumference, so that E the centre of the circle be without the angle BDC, and join DE, and produce it to C. 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. XIII. 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 contre, B 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 (12. 3.)

A

F

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.

F

But, if the segment BAD 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 B it to C, and join CE: 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.

COR. Segments of circles which contain equal angles are similar.

PROP. XIV. THEOR.

The opposite angles of any quadrilateral figure described 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.

D

Join AC, BD; and because the three angles of every triangle are equal (5. 2.) to two right angles, the three angles of the triangle CAB, viz. the angles CAB, ABC, BCA are equal to two right angles: but the angle CAB is equal (13. 3.) 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

:

A

B

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. XV. THEOR.

Upon the same straight line, and upon the same side of it, there cannot be two similar segments of circles, not coinciding with 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 coin