The Elements of Euclid; viz. the first six books, together with the eleventh and twelfth. Also the book of Euclid's Data. By R. Simson. To which is added, A treatise on the construction of the trigonometrical canon [by J. Christison] and A concise account of logarithms [by A. Robertson].

the angle C is equal to the angle in the segment AHB. † 1 Ax. 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.

PROB.—From a given circle to cut off a segment, 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 from the circle ABC, a segment that shall contain an angle equal to the given angle D. Draw the straight line EF touching the circle ABC in the 17.3. point B; and at the point B, in the straight line BF, make the angle FBC

equal to the angle D: the segment BAC shall contain an angle equal to the given D angle D.

Because the straight line EF touches the circle ABC, and BC is drawn from

23. 1.

the point of contact B, the angle FBC is equal to the angle 32. 3. in the alternate segment BAC of the circle: but the angle FBC is equal to the angle D; therefore the angle in the † Constr. segment BAC is equal to the angle D. Wherefore, from the given circle ABC, the segment BAC is cut off, containing an angle equal to the given angle D. Which was to be done.

PROPOSITION XXXV.

1 Ax.

THEOR.-If two straight lines cut one another within a circle, See N. the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the olher.

Let the two straight lines AC, BD cut one another in the point E, within the circle ABCD: the rectangle contained by AE, EC, shall be equal to the rect- A

angle 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 † 15 Def. 1.

likewise equal to the rectangle BE, ED.

3. 3.

* 5.2.

* 47. 1.

†1 Ax. † 3 Ax.

• 12. 1. * 3. 3.

* 5.2.

† 2 Ax.

* 47. 1.

5. 2.

† 1 Ax.

† 3 Ax.

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 is equal to EC: and because
the straight line BD is cut into two equal parts
in the point F, and into two unequal parts in
the point E, the rectangle BE, ED, together
with the square of EF, is equal * to the square
of FB; that is, to the square of FA: but the
squares of AE, EF are equal to the square 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 * FG perpendicular to AC; therefore AG is equal to GC; wherefore the rectangle AE, EC, together with the square of EG, is equal 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 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 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.

Because AE has been proved to be equal to EC.

Lastly, let neither of the straight lines AC, BD pass through the centre: take † the centre F, and through E, the intersec- † 1. 3. tion of the straight lines AC, DB, draw the diameter GEFH and because the rectangle AE, EC is equal, as has been shewn, 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 † 1 Ax. two straight lines, &c. Q. E. D.

PROPOSITION 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 let DCA, DB be two straight lines drawn from it, of which DCA cuts the circle, and DB touches the same: the rectangle AD, DC shall be equal to the square of DB.

Ꭰ

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 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 to the square B of ED: but 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

⚫ 18. 3.

6. 2.

47. 1.

cause EBD is a right angle therefore the rectangle AD, DC, together with the square of EB, is equal to the squares of †1 Ax. EB, BD: take away the common square of EB; therefore the remaining rectangle AD, DC is equal to the square of the † 3 Ax. tangent DB.

But if DCA does not pass through the centre of the circle ABC, take the centre E, and draw EF perpendicular to 1.3. AC, and join EB, EC, ED. And because the straight line EF which passes through the centre, cuts the straight line AC

12. 1.

8. 3.

* 6.2.

† 2 Ax.

* 47. 1.

† 1 Ax.

* 47. 1.

†3 Ax.

See N.

square

which does not pass through the centre, at right angles, it
also bisects it; therefore AF is equal to FC: and because
the straight line AC is bisected in F, and pro-
duced to D, the rectangle AD, DC, together
with the square of FC, is equal
to the square
of FD: to each of these equals add the
of FE; therefore the rectangle AD, DC, to-
gether with the squares of CF, FE, is equal † to
the squares of DF, FE: but the square of ED
is equal to the squares of DF, FE, because EFD is a right
angle; and for the same reason, 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 :
but CE is equal to EB; therefore the rectangle AD, DC, to-
gether with the square of EB, is equal to the square of ED:
but the squares of EB, BD are equal to the square
* 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 re-
maining rectangle AD, DC is equal to the square of DB.
Wherefore, if from any point, &c. Q. E. D.

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 without the
circle, are equal to one another; viz. the rect-
angle 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.

PROPOSITION 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 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 shall touch the circle.

* 17. 3.

1.3.

18. 3.

Draw the straight line DE, touching the circle ABC; find↑ its centre F, and join FE, FB, FD; then FED is a right* angle: and because DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal to the square of DE: but the rect- 36. 3. angle 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 are equal to DB, BF, each to each; and the base FD is common to the two triangles DEF, DBF ; therefore the angle DEF is equal to the angle DBF:

† 1 Ax.

† 15 Def. 1.

but DEF was shewn to be a right angle; therefore also DBF

8. 1.

is a right angle: and BF, if produced, is a diameter; and † 1 Ax. the straight line which is drawn at right angles to a diameter,

from the extremity of it, touches the circle: therefore DB Cor. 16.3. touches the circle ABC. Wherefore, if from a point, &c. Q. E. D.

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The Elements of Euclid; viz. the first six books, together with the eleventh ...