ADC is greater than a right angle. Therefore, in a circle, Book III. &c. Q. E. D.

COR. From the demonstration it is manifest that if one angle of a triangle be equal to the other two angles, it is a right angle.

PROP. XXXII. THEOR.

IF a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line which touches the circle will be equal to the angles in the alternate segments of the circle.

Let the straight line EF touch the circle ABCD in B, and from the point B let the straight line BD be drawn cutting the circle; the angles which BD makes with the touching line EF will be equal to the angles in the alternate segments of the circle; that is, the angle FBD is equal to the angle which is in the segment DAB, and the angle DBE to the angle in the segment BCD.

A

b 19.3.

c 31. 3.

d 32. 1.

From the point B drawa BA at right angles to EF, and a 11. 1. take any point C in the arch BD, and join AD, DC, CB. Because the straight line EF touches the circle ABCD in the point B, and BA is drawn at right angles to the touching line, from the point of contact B, the centre of the circle isb in BA; therefore the angle ADB in a semicircle is a right angle, and consequently the other two angles BAD, ABD are equald to a right angle. But ABF is likewise a right angle; therefore the angle ABF is equal to the angles BAD, ABD. Take from these equals the common

E

B

F

angle ABD, and there will remain the angle DBF equal to the angle BAD, which is in the alternate segment of the

e 22. 3. f 13. 1.

Book III. circle. Again, because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are equale to two right angles; therefore the angles DBF, DBE, being likewise equal to two right angles, are equal to the angles BAD, BCD. But DBF has been proved equal to BAD; therefore the remaining angle DBE is equal to the angle BCD in the alternate segment of the circle. Wherefore, if a straight line, &c. Q. E. D.

a 10. 1.

b 31. 3.

PROP. XXXIII. PROB.

UPON a given straight line to describe a segment of a circle, containing an angle equal to a given rectilineal angle.

Let AB be the given straight line, and the angle at C the given rectilineal angle; it is required to describe upon the given straight line AB a segment of a circle, containing an angle equal to the angle C.

First, let the angle at C be a right angle; bisecta AB in F, and from the centre F, at the distance FB, describe the semicircle AHB; the angle AHB in a semicircle. is equal to the right angle at

C.

If the angle C be not a right angle, at the point A, in

the straight line AB, make the angle BAD equal to the an- Book III. gle C, and from the point A drawd AE at right angles to AD; bisecta AB in F, and from F drawd FG at right an- c23.1. gles to AB, and join GB. Because AF is equal to FB, and d 11. 1. FG common to the triangles AFG, BFG, the two sides AF, FG are equal to the two BF, FG; also the angle AFG is equal to the angle BFG; therefore the base AG is equale to e 4. 1. the base GB; therefore the circle described from the centre G, at the distance GA, will pass through the point B. Let this be the circle AHB. Then, because from the point A, the extremity of the diameter AE, AD is drawn at right angles to AE, AD touches the circle; and because AB, fCor. 16.3. drawn from the point of contact A, cuts the circle, the angle DAB is equal to the angle in the alternate segment AHB. g 32. 3. But the angle DAB is equal to the angle C, therefore also the angle C is equal to the angle in the segment AHB. Wherefore, upon the given straight line AB the segment AHB of a circle is described which contains an angle equal to the given angle at C. Which was to be done.

PROP. XXXIV. PROB.

TO cut off a segment from a given circle which shall contain an angle equal to a given rectilineal angle.

Let ABC be the given circle, and D the given rectilineal angle; it is required to cut off a segment from the circle ABC that shall contain an angle equal to the angle D.

Draw the straight line EF touching the circle ABC in a 17. 3. the point B, and at the

Book III. ment BAC. But the angle FBC is equal to the angle D; therefore the angle in the segment BAC is equal to the angle D. Wherefore, the segment BAC is cut off from the given circle ABC containing an angle equal to the given angle D. Which was to be done.

a 3. 3.

b 5.2.

c 47. 1.

PROP. XXXV. THEOR.

IF two straight lines within a circle cut each other, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other.

Let the two straight lines AC, BD, within the circle ABCD, cut each other in the point E; the rectangle contained by AE, EC is equal to the rectangle contained by BE, ED.

If AC, BD pass through the centre, so that E is the centre, it is evident that AE, EC, BE, ED, being all equal, the rectangle AE.EC is equal to the rectangle

BE.ED.

A

E

B

D

D

Let one of them BD pass through the centre, and cut the other AC, which does not pass through the centre, at right angles in the point E; bisect BD in F, then F is the centre of the circle ABCD. Join AF. Because BD, which passes through the centre, cuts, the straight line AC, which does not pass through the centre, at right angles in E, AE, EC are equal to each other; and because the straight line BD is cut into two equal parts in the point F, and into two unequal in the point E, BE.ED + EF2 : FB2-AF2. But AF2-AE2+cEF2; therefore

=

F

A

E

B

BE,ED+EF2=AE2+EF2; therefore if EF2 be taken from

each, BE.ED-AE2-AE.EC.

D

d 12. 1.

Let BD, which passes through the centre, cut the other Book III AC, which does not pass through the centre, in E, but not at right angles; bisect BD in F, then F is the centre of the circle. Join AF, and from F drawd FG perpendicular to AC; therefore AG is equal to GC; wherefore AE.EC+EG-AG2; therefore if GF2 be added to both, AE.EC+EG2+GF2=AG2+GF2. Now EG+GF2=EF2, and AG2 +GF2-AF2; therefore AE.EC +EF2-AF2-FB2.

E

C

G

B

But FB2-BE.ED+EF2; therefore

AE.EC+EF2-BE.ED+EF2; therefore, if EF2 be taken

PROP. XXXVI. THEOR.

IF from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it, the rectangle contained by the whole line which cuts the circle and the part of it without the circle is equal to the square of the line which touches it.

Let D be any point without the circle ABC, and DCA, DB two straight lines drawn from it, of which DCA cuts