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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, so make DB to BA, and complete the parallelogram AC: AC is equal to the rectangle FH, GK: and the square of

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BC 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, BG, that is, the parallelogram AC, is equal to the rectangle FH, GK.

PROP. XCI.

Ir a straight line drawn within a circle given in magnitude cuts 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.

88.

Take D the centre of the circle, join AD, and produce 1. 3.

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it to E, and join EC: The angle ACE 31. 3. being a right angle, is given; and the 43 Dat, 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 5 Def, the circle is given in magnitude: AC 2 Dat is therefore given in magnitude.

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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.

Take D the centre of the circle, join AD and produce it to E, and join EC: And because each of the straight lines EA and AC is given, their ratio is 1 Dat. given a; and the angle ACE is a right A angle, therefore, the triangle ACE is 46 Dat. given in species, and consequently the angle AEC is given.

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B

E

D

IF from any point in the circumference of a circle given in 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.

A

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 B join, BD, DC; and because each of the

29 Dat, points B, D, is given, BD is givena in

position; and because the angle BAC is given, the angle

20. 3. BDC is given, therefore because the straight line DC is

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drawn to the given point D, in the straight line BD given in position in the given angle BDC, DC is given in posi- 32 Dat. tion: And the circumference ABC is given in position, therefored the point C is given.

PROP. 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.

B

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 givena in position and magni-C tude: And DBA 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 givend in position, there

d 28 Dat.

91.

29 Dat. b 18. 3.

Cor. 5. 4.

d 6 Def.

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fore the point B is given. The point A is also given: 28 Dat. Therefore the straight line AB is givena in position and magnitude.

PROP. XCV.

Ir 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.

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.

From the point A drawa AD touching the circle; therefore ADC is given in position and magnitude: And because AD is given, the square of AD is given, which is

92.

D

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56 Dat.

equal to the rectangle BA, AC: Therefore the rectangle a 36. 3.

BA, AC, is given.

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Ir 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.

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 * 29 Dat. straight line AD is given a in position; and the circle BEC is given in position; 28 Dat, therefore the points E, F, are givenb; and the point A is given, therefore EA,

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D

A

E

AF, are each of them given a, and the rectangle EA, AF, 35. 3. is therefore given; and it is equal to the rectangle BA, AC; which consequently is given.

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PROP. XCVII.

Ir 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 rectangle contained by both these lines together which contain the given angle, and part of the bisecting line cut off below 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.

Join BD; and because BC is drawn within the circle ABC

C

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given in magnitude cutting off the segment BAC, containing the given angle BAC; BC is given" in magnitude: By a 91 Dat the same reason BD is given; therefore the ratio of BC 1 Dat. to BD is given: And because the angle BAC is bisected by AD, as BA to AC, so is BE to EC; and, by permutation, 3. 6. as AB to BE, so is AC to CE; wherefore, as BA and AC 12. 5. together to BC, so is AC to CE; And because the angle BAE is equal to EAC, and the angle ACE to ADBe, the triangle ACE is equiangular to the triangle ADB; therefore as AC to CE, so is AD to DB: But as AC to CE, so is BA together with AC to BC; as therefore BA and AC to BC, so is AD to DB; and by permutation,

E

B

as BA and AC to AD, so is BC to BD: And the ratio of BC to BD is given, therefore the ratio of BA together with AC to AD is given.

Also the rectangle contained by BA and AC together, and DE, is given.

Because the triangle BDE is equiangular to the triangle ACE, as BD to DE, so is AC to CE; and as AC to CE, so is BA and AC to BC; therefore as BA and AC to BC, so is BD to DE; wherefore the rectangle contained by BA and AC together, and DE, is equal to the rectangle CB, BD: But CB, BD, is given; therefore the rectangle contained by BA and AC together, and DE, is given.

Otherwise,

21. 3.

Produce CA, and make AF equal to AB, and join BF; and because the angle BAC is double of each of the angles. $5. & BFA, BAD; the angle BFA is equal to BAD; and the angle BCA is equal to BDA, therefore the triangle FCB is equiangular to ABD: As therefore FC to CB, so is AD to DB; and, by permutation, as FC, that is, BA and AC together, to AD, so is CB to BD: And the ratio of CB to BD is given, therefore the ratio of BA and AC to AD is given.

And because the angle BFC is equal to the angle DAC, that is, to the angle DBC, and the angle ACB equal to the angle ADB; the triangle FCB is equiangular to BDE: as therefore FC to CB, so is BD to DE; therefore the rectangle contained by FC, that is, BA and AC together, and

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