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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 C

H н to the angle in the alternate seg

F ment AHB (32. 3.): but the an A

B gle 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.

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PROP. XXXIV. PROB.

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

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Let ABC be the given circle, and D the given rectlineal 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 point B, and at the point

А 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 D contact B, the angle FBC is equal (32. 3.) to the angle in the alternate segment BAC of the

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B F circle : 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.

PROP. XXXV. THEOR.

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If 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 A E D 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 B rectangle BE, ED.

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

D 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

F point E, the rectangle BE, ED together with the square of EF, is equal

A (5. 2.) to the square of FB ; that is, to

E the square of FA; but the squares of AE, EF are equal (47. 1.) to the square

B of FA ; therefore the rectangle BE, ED, together with the square of EF, is equal to the squares of AE, EF : take away the common square of EF, and the remaining rectangle BE, ED is equal to the remaining square of AE; that is, to the rectangle AE, EC.

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

* See Note

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perpendicular to AC; therefore AG is equal (3. 3.) to GC; 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

D 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 AĞ, GF are equal to the square of AF: therefore the rectangle A

с AE, EC, together with the square

G EF, is equal to the square of AF; that

B is, to the square of FB: but the square of FB is equal (5. 2.) to 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, ED, 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.

Lastly, Let neither of the straight lines AC, BD pass through the centre: take the centre F, and through н E, the intersection of the straight lines AC, DB draw the diameter GEFH: and

F

ID because the rectangle AE, EC is equal, as has been shown to the rectangle GE,

E EH : and, for the same reason, the rect- A

C angle BE, ED is equal to the same rectangle GE, EH; therefore the rectangle

G AE, EC is equal to the rectangle BE,

B ED. Wherefore, if two straight lines, &c. Q. E. D.

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, 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 cir

cle, and DB touches the same; the rectangle AD, DC is equal to the square of DB.

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

D because the straight line AC is bisected in E, and produced to the point D, the rectangle AD, DC, together with the square of

С EC, is equal (6. 2.) to the square of ED, and CE is equal to EB: therefore the rect

В. angle 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

E 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

A square of EB; therefore the remaining rectangle AD, DC is equal to the square of the tangent DB.

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 D 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

B add the square of FE; therefore the rect

T angle AD, DC, together with the squares of

E 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 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,

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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 Cor. If from any point without a circle, there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of them with

F out the circle, are equal to one another, viz. the rectangle BA, AE to the rectangle CA, AF : for each of them is equal to the square of the straight line AD which touches the circle.

B

PROP. XXXVII. THEOR.

If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it; if the rectangle contained by the whole line, which cuts the circle, and the part of it without the circle be equal to the square of the line which meets it, the line which meets it shall touch the circle. *

Let any point D be taken without the circle ABC, and from it let two straight lines DCA and DB be drawn, of which DCA cuts the circle, and DB meets it, if the rectangle AD, DC be equal to the square of DB; DB touches the circle.

Draw (17. 3.) the straight line DE touching the circle ABC, find its centre F, and join FE, FB, FD; then FED is a right (18. 3.) angle : and because DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal (36. 3.) to the square of DE: but the rectangle AD, DC is, by hypothesis, equal to the square of DB : therefore the square of DE is equal to the square of DB; and the straight line DE equal to the straight line DB ; and FE is equal to FB, wherefore DE, EF

See Note.

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