PROP. XXXII. THEOR.

Ir 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 touching the circle, shall be equal to the angles which are 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 DB makes with the touching line EF shall 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

D

C

From the point B draw (11. 1.) BA at right angles to EF, and take any point C in the circumference BD; and join AD, DC, CB; and 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 is (19. 3.) in BA; therefore the angle ADB in a semicircle is a right (31. 3.) angle, and consequently the other two angles BAD, ABD are equal (32. 1.) to a right angle: but ABF is likewise a right angle; therefore the angle ABF is equal to the angles BAD, E ABD: take from these equals the common angle ABD; therefore the remaining angle DBF is equal to the angle BAD, which is in the alternate segment of the circle: and because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are equal (22. 3.) to two right angles; therefore the angles DBF, DBE, being likewise equal (13. 1.) to two right angles, are equal to the angles BAD, BCD; and 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.

PROP. XXXIII. PROB.

B

F

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.

* See Note.

H

First, let the angle at C be C a right angle, and bisect (10. 1.) AB in F, and from the centre F, at the distance FB, describe the semicircle AHB; therefore the angle AHB in a semicircle is (31. 3.) equal to

the right angle at C.

But, if the angle C be not a right angle, at the point A, in the straight line AB, make (23. 1.) the angle BAD equal to the angle C,

and from the point A draw (11. 1.) AE at right angles to AD: bisect (10. 1.) AB in F, and from F draw (11. 1.) FG at right angles to AB, and join GB: and because AF is equal to FB, and FG common to the triangles AFG, BFG, the two sides AF, FG are equal to the two BF, FG; and the angle AFG is equal to the angle BFG; therefore the base AG is equal (4. 1.) to the base GB; and the circle described

D\

H

G

E

B

F

C

F

A

B

from the centre G, at the distance GA, shall pass through the point B; let this be the circle AHB: and because from the point A the extremity of the diameter AE, AD is drawn at right angles to AE, therefore AD (Cor. 16. 3.) touches the circle; and because AB drawn from the point of contact A cuts the circle, the angle DAB is equal to the angle in the alternate segment AHB (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.

D

PROP. XXXIV. PROB.

E

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 given angle D.

Draw (17. 3) the straight line EF touching the circle ABC in the

A

C

point B, and at the point B, in the straight line BF, make (23. 1.) the angle FBC equal to the angle D; therefore, because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, the angle FBC is equal (32. D' 3.) to the angle in the alternate segment BAC of the circle: but the angle FBC is equal to the angle D; E 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. done.

B

F

Which was to be

PROP. XXXV. THEOR.

Ir two straight lines within a circle cut one another, 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 one another in the point E: the rectangle contained by AE, EC is equal to the rectangle contained by BE, ED.

If AC, BD pass each of them 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 likewise equal to the rectangle BE, ED.

A

E

D

B

D

C

But 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; then, if BD be bisected in F, F is the centre of the circle ABCD; join AF: and 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 (3. 3.) to one another : and because the straight line BD is cut into two equal parts in the point F, and into two unequal in the point E, the rectangle BE ED, together with the square of BF, is equal (5. 2.) to the square FB; that is, to the square of A FA; but the squares of AE, EF are equal (47. 1.) to the square of FA; therefore the rectangle BE, ED, to

F

C

E

B

gether with the square of EF, is equal to the squares of AE, EF: take away the common square of EF, and the remaining rectangle

* See Note.

BE, ED is equal to the remaining square of AE; that is, to the rectangle AE, EC.

D

Next, let BD, which passes through the centre, cut the other AC, which does not pass through the centre, in E, but not at right angles: then, as before, if BD be bisected in F, F is the centre of the circle. Join AF, and from F draw (12. 1.) FG perpendicular to AC; therefore AG is equal (3. 3.) to GD; wherefore the rectangle AE, EC, together with the square of EG, is equal (5. 2.) to the square of AG: to each of these equals add the square of GF; therefore the rectangle AE, EC, together with the squares of EG, GF, is equal to the squares of AG, GF: but the squares of EG, GF are equal (47. 1.) to the square of EF, and the squares of AG, GF are equal to the square of AF; therefore the rectangle AE, EC, together with the square of EF, is equal to the square of AF; that is, to the square of FB: but the square of FB is equal (5. 2.) to A the rectangle BE, ED, together with the square of EF: therefore the rectangle AE, EC, together with the square of EF, is equal to the rectangle BE, DE, together with the square of EF: take away the common square of EF, and the remaining rectangle AE, EC is therefore equal to the remaining rectangle BE, ED.

F

E

C

G

B

Lastly, let neither of the straight lines AC, BD pass through the centre: take the centre F, and through E, the intersection of the straight lines AC, DB draw the diameter GEFH: and because the rectangle AE, EC is equal, as has been shown, to the rectangle GE, EH: and, for the same reason, the rectangle BE, ED is equal to the same rectangle GE, EH; therefore the rectangle AE, EC is equal to the rectangle BE, ED. Wherefore, if two straight lines, &c. Q. E. D.

A

PROP. XXXVI. THEOR.

F

D

F

C

G

B

Ir 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, shall be 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 the circle, and DB touches the same; the rectangle AD, DC is equal to the square of DB.

D

C

Either DCA passes through the centre, or it does not; first, let it pass through the centre E, and join EB; therefore the angle EBD is a right (18. 3.) angle: and because the straight line AC is bisected in E, and produced to the point D, the rectangle AD, DC, together with the square of EC, is equal (6. 2.) to the square of ED, and CE is equal to EB: therefore the rectangle AD, DC, together with the square of EB, is equal to the square of ED: but the square of ED is equal (47. 1.) to the squares of EB, BD because EBD is a right angle: therefore the rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD: take away the common square of EB; therefore the remaining rectangle AD, DC is equal to the square of the tangent DB.

В

B

F

A

E

Ꭰ

But if DCA does not pass through the centre of the circle ABC, take (1. 3.) the centre E, and draw EF perpendicular (12. 1.) to AC, and join EB, EC, ED: and because the straight line EF, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles, it shall likewise bisect (3. 3.) it; therefore AF is equal to FC: and because the straight line AC is bisected in F, and produced to D, the rectangle AD, DC, together with the square of FC, is equal (6. 2.) to the square of FD: to each of these equals add the square of FE; therefore the rectangle AD, DC, together with the squares of CF, FE, is equal to the squares of DF, FE: but the square of ED is equal (47. 1.) to the squares of DF, FE, because EFD is a right angle: and the square of EC is equal to the squares of CF, FE; therefore the rectangle AD, DC, together with the square of EC, is equal to the square of ED: and CE is equal to EB; therefore the rectangle AD, DC, together with the square of EB, is equal to the square of ED: but the squares of EB, BD are equal to the square (47. 1.) of ED, because EBD is a right angle; therefore the rectangle AD, DC, together with the square of EB, is equal to the squares of EB, BD: take away the common square of EB: therefore the remaining rectangle AD, DC is equal to the square of DB. Wherefore, if from any point, &c. Q. E. D.

A

E