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*22.6.

1.6.

14.5.

remainder shall have to the square of BA the same ratio which the square of NH has to the square of HG.

Because, by the construction, the square of BC is equal to the square of BF together with the rectangle LH, HM; if from the square of BC there be taken the rectangle LH, HM, there remains the square of BF, which has * to the square of BA the same ratio which the square of NH has to the square of HG, because, as NH to HG, so FB was made to BA; but as HG to GK, so is BA to AE, because the triangle GHK is equiangular to ABE; therefore, ex æquali, as NH to GK, s0 is FB to AE; wherefore the rectangle NH, HL is to the rectangle GK, HL, as the rectangle FB, BC to AE, BC; but, by the construction, the rectangle NH, HL is equal to FB, BC; therefore the rectangle GK, HL is equal to the rect angle AE, BC, that is, to the parallelogram AC.

The analysis of this problem might have been made as in the 86th Prop. in the Greek, and the composition of it may be made as that which is in Prop. 87th of this edition.

PROPOSITION XC.

(0.)

If two straight lines contain a given parallelogram in a given angle, and if the square of one of them, together with the space which has a given ratio to the square of the other, be given, each of the straight lines shall be given.

Let the two straight lines AB, BC contain the given parallelogram AC in the given angle ABC, and let the square of BC together with the space which has a given ratio to the square of AB be given: AB, BC are each of them given.

• 43 Dat.

Let the square of BD be the space which has the given ratio to the square of AB: therefore, by the hypothesis, the square of BC together with the square of BD is given. From the point A, draw AE perpendicular to BC; and because the angles ABE, BEA are given, the triangle ABE is given * in species: therefore the ratio of BA to AE is given: and because the ratio of the square of BD to the

58 Dat.

*9 Dat.

square of BA is given, the ratio of the straight
line BD to BA is given *; and the ratio of BA
to AE is given; therefore the ratio of AE to
BD is given, as also the ratio of the rectangle AE,
BC, that is, of the parallelogram AC, to the rectangle DB,

D

A

BE

C

BC; and AC is given, therefore the rectangle DB, BC is given;. and the square of BC together with the square of BD is given: therefore because the rectangle contained by the two straight lines DB, BC is given, and the sum of their squares is given, the straight lines DB, BC are each of them given; and the ratio of DB to BA is given; therefore AB, BC are given.

The Composition is as follows :

Let FGH be the given angle to which the angle of the parallelogram is to be made equal, and from any point Fin GF, draw FH perpendicular to GH; and let the rectangle FH, GH be that to which the parallelogram is to be made equal; and let the rectangle KG, GL be the space to which the square of one of the sides of the parallelogram, together with the space which has a given ratio to the square of the other side is to be made equal; and let this given ratio be the same which the square of the given straight line MG has to the square of GF.

By the 88th dat. find two straight lines DB, BC, which contain a rectangle equal to the given rectangle MG, GK, and such that the sum of their squares is equal to the given rectangle KG, GL; therefore, by the determination of the problem in that proposition, twice the rectangle MG, GK must not be greater than the rectangle KG, GL. Let it be so, and join the straight lines DB, BC, in the angle DBC, equal to the given angle FGH; and, as MG to GF,

68 Dat.

[blocks in formation]

together with the square of BD, which, by the construction, has to the square of BA the given ratio which the square of MG has to the square of GF, is equal, by the construction, to the given rectangle KG, GL. Draw AE perpendicular to

BC.

Because, as DB to BA, so is MG to GF; and as BA to AE, so GF to FH; ex æquali, as DB to AE, so is MG to FH; therefore as the rectangle DB, BC to AE, BC, so is the rectangle MG, GK to FH, GK; and the rectangle DB, BC is equal to the rectangle MG, GK; therefore the rectangle AE, BC, that is, the parallelogram AC, is equal to the rectangle FH, GK.

DD

(88.)

1.3.

31.3.

* 43 Dat.

* 5 Def.

* 2 Dat.

(89.)

PROPOSITION XCI.

If a straight line drawn within a circle given in magnitude, cut off a segment which contains a given angle, the straight line is given in magnitude.

In the circle ABC given in magnitude, let the straight line AC be drawn, cutting off the segment AEC which contains the given angle AEC: the straight line AC is given in magnitude.

B

E

D

C

Take D the centre of the circle *, join AD, and produce it to E, and join EC: the angle ACE being a right* angle, is given; and the angle AEC is given; therefore the triangle ACE is given in species, and the ratio of EA to AC is therefore given, and EA is given in magnitude, because the circle is given * in magnitude: AC is therefore given * in magnitude.

PROPOSITION XCII.

A

CAB

300

[blocks in formation]

If a straight line given in magnitude be drawn within a circle given in magnitude, it shall cut off a segment containing a given angle.

Let the straight line AC given in magnitude be drawn within the circle ABC given in magnitude: it shall cut off a segment containing a given angle.

B

E

Take D the centre of the circle, join AD

D

* 1 Dat.

* 46 Dat.

and produce it to E, and join EC: and because
each of the straight lines EA and AC is given,
their ratio is given *: and the angle ACE is a
right angle, therefore the triangle ACE is given * in species.
and consequently the angle AEC is given.

[blocks in formation]

(90.)

PROPOSITION XCIII.

If from any point in the circumference of a circle given is position, two straight lines be drawn, meeting the circumference and containing a given angle; if the point in which one of them meets the circumference again be given, the point in which the other meets it is also given.

From any point A, in the circumference of a circle ABC given in position, let AB, AC be drawn to the circumference, making the given angle BAC: if the point B be given, the point C is also given.

Take D the centre of the circle, and join BD, DC, and because each of the points B, D is given, BD is given in position; and because the angle BAC is given, the angle BDC is given *, therefore because the straight line DC is drawn to

the given point D, in the straight line BD

A

29 Dat.

* 20.3.

D

B

C

given in position, in the given angle BDC, DC
is given * in position. And the circumference ABC is given

in position, therefore the point C is given.

PROPOSITION XCIV.

If from a given point a straight line be drawn touching a circle given in position, the straight line is given in position and magnitude.

Let the straight line AB be drawn from the given point A, touching the circle BC given in position: AB is given in position and magnitude.

Take D the centre of the circle, and join DA, DB: Because each of the points D, A is given, the straight line AD is given * in position and magnitude: and BDA is a right * angle, wherefore DA is a diameter * of the circle DBA described about the triangle DBA: and that circle is therefore given in position: and the circle BC is given * in position, therefore the point B is given *. The point A is also given: therefore the straight line AB is given * in position and magnitude.

PROPOSITION XCV.

32 Dat.

28 Dat.

(91.)

29 Dat.

* 18.3.

C

B

* Cor. 5. 4.

A

* 6 Def. 28 Dat.

29 Dat.

If a straight line be drawn from a given point without a circle given in position, the rectangle contained by the segments betwixt the point and the circumference of the circle is given.

(92.)

Let the straight line ABC be drawn from the given point A without the circle BCD given in position, cutting it in B, C: the rectangle BA, AC is given.

D

C

BA

From the point A, draw * AD touching the circle; wherefore AD is given * in position and magni

* 17.3.

94 Dat.

* 56 Dat.

* 36. 3.

(93.)

* 29 Dat.

* 28 Dat.

* 29 Dat.

* 35. 3.

tude: and because AD is given, the square of AD is given *, which is equal * to the rectangle BA, AC: therefore the rectangle BA, AC is given.

PROPOSITION XCVI.

If a straight line be drawn through a given point within a circle given in position, the rectangle contained by the segments betwixt the point and the circumference of the circle is given.

Let the straight line BAC be drawn through the given point A, within the circle BCE, given in position: the rectangle BA, AC is given.

B

F

E

D

A

C

Take D, the centre of the circle, join AD, and produce it to the points E, F: because the points A, D are given, the straight line AD is given * in position; and the circle BEC is given in position; therefore the points E, F are given *; and the point A is given, therefore EA, AF are each of them given *, and the rectangle EA, AF is therefore given; and it is equal * to the rectangle BA, AC; which consequently is given.

(94.)

PROPOSITION XCVII.

If a straight line be drawn within a circle given in magnitude, cutting off a segment containing a given angle; if the angle in the segment be bisected by a straight line produced till it meets the circumference, the straight lines which contain the given angle shall both of them together have a given ratio to the straight line which bisects the angle: and the rectangk contained by both these lines together which contain the gives angle, and the part of the bisecting lines cut off belore the base of the segment, shall be given.

Let the straight line BC be drawn within the circle ABC given in magnitude, cutting off a segment containing the given angle BAC, and let the angle BAC be bisected by the straight line AD: BA together with AC has a given ratio to AD; and the rectangle contained by BA and AC together, and the straight line ED cut off from AB below BC the base of the segment, is given.

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