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because the straight line DC is drawn to the given point D in the straight line BD given in position in the given angle BDC, DC is given c in position : -and the circumference ABC is given c 32. dar. in position, therefored the point C is given.

d 28. dar,

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.

Take D the centre of the circle, and join DA, DB : because each of the points D, A is given, the

B straight line AD is given a in position

a 29. da. and magnitude: and DBA is a rightb

b 18. 3. angle, wherefore DA is a diameter of the circle DBA, described about the tri

c Cor.5.4 D

А angled DBA; and that circle is there

d 6. defa fore given d in position: and the circle BC is given in position, therefore the point B is givene ; the point A is also given; therefore the e 28. dat. straight line AB is given a in position and magnitude.

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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.

Let the straight line ABC be drawn from the given point A without the circle BCD given in posi

D tion, cutting it in B, C; the rectangle BA, AC is given.

2 17.3. From the point A draw a AD touch- Ç

BA b 94. dat, ing the circle; therefore AD is given b in position and magnitude; and because AD is given, the square of AD is given', which is equal to the rectangle BA, AC; therefore the 56. daß

d 36.3: Hectangle BA, AC is given.

PROP. 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.
Take D the centre of the circle, join AD,

E
and produce it to the points E, F; because

the points A, D are given, the straight lino a 29. dat. AD is givena in position ; and the circle BEC

is given in position ; therefore the points B b 28. dat. E, F are given b; and the point A is given,

A therefore EA, AF are each of them given a ; c 35. 3. and the rectangle EA, AF is therefore given ; and it is equal to

the rectangle BA, AC, which consequently is given,

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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 rectangle contained by both these lines together which contain the given angle, and the

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 gi. ven 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 AD below BC the base of the segment, is given.

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

given in magnitude cutting off the segment BAC, containing the given angle BAC; BC is given a in magnitude ; by the same a 91. da:. reason BD is given ; therefore b the ratio of BC to BD is given : b 1. dat. and because the angle B.4C is bisected by AD, as BA to AC, so is © BE to EC ; and, by permutation, as AB to BE, so is AC c 3. 6. to CE: wherefore d as BA and AC together to BC, so is AC tod 12. 3. CE:and because the angle BAE is equal to EAC, and the angle ACE to e ADB, the triangle ACE is F

e 21.3. equiangular to the triangle ADB;

A 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

E to DB; and, by permutation, as BA

B

C. 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 10 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,

55 & 2

32.1.

Produce CA, and make AF equal to AB, and join BF; and because the angle BAC is double a of each of the angles 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 per. mutation, 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 tlie, 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 DE is equal to the

rectangle CB, BD, which is given, and therefore the rectangle contained by BA, AC together, and DE is given.

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IF a straight line be drawn within a circle given in magnitude, cutting off a segment containing a given angle : if the angle adjacent to the angle in the segment be bisected by a straight line produced till it meet the circumference again and the base of the segment; the excess of the straight lines which contain the given angle shall have a given ratio to the segment of the bisect. ing line which is within the circle ; and the rectangle contained by the same excess and the segment of the bisecting line betwixt the base produced and the point where it again meets the circumference, 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 CAF adjacent to BAC be bisected by the straight line Dal meeting the circumference again in D, and BC the base of the segment produced in E; the excess of BA, AC has a given ratio to AD; and the rectangle which is contained by the same excess and the straight line ED, is given.

Join BD, and through B draw BG parallel to De meeting AC produced in G: and because BC cuts off from the circle ABC given in magnitude the seg

D

F ment BAC containing a given ana 91. dat. gle, BC is therefore given a in

magnitude : by the same reason
BD is given, because the angle
BAD is equal to the given angle

E.
EAF: therefore the ratio of BC to B

С
BD is given: and because the an-
gle CAE is equal to EAF, of

G
which CAE is equal to the alternate angle AGB, and EAF to
the interior and opposite angle ABG; therefore the angle AGB
is equal to ABG, and the straight line AB equal to AG; so that

GC is the excess of BA, AC; and because the angle BGC is equal to GAE, that is, to EAF, or the angle BAD; and that the angle BCG is equal to the opposite interior angle BDA of the quadrilateral BCAD in the circle ; therefore the triangle BGC is equiangular to DA : therefore as GC to CB, so is AD to DB; and, by permutation, as GC which is the excess of BA, AC to AD, so is CB to BD: and the ratio of CB to BD is given : therefore the ratio of the excess of BA, AC to AD is given.

And because the angle GBC is equal to the alternate angle DEB, and the angle BCG equal to BDE; the triangle BCG is equiangular lo BDE: therefore as GC to CB, so is BD to DE; and consequently the rectangle GC, DE is equal to the rectangle CB, BD which is given, because its sides CB, BD are given: therefore the rectangle contained by the excess of BA, AC and the straight line DE is given.

PROP. XCIX.

95.

IF from a given point in the diameter of a circle given in position, or in the diameter produced, a straight line be drawn to any point in the circumference, and from that point a straight line be drawn at right angles to the first, and from the point in which this meets the circumference again, a straight line be drawn parallel to the first; the point in which this parallel meets the diameter is given; and the rectangle contained by the two parallels is given.

In BC the diameter of the circle ABC given in position, or in BC produced, let the given point D be taken, and from D let a straight line DA be drawn to any point A in the circumference, and let AE be drawn at right angles lo DA, and from the point E where it meets the circumference again let EF be drawn parallel to DA meeting BC in F; the point F is given, as also the rectangle AD, EF.

Produce EF to the circumference in G, and join AG: because GEA is a right angle, the straight line AG is a the dia- a Cor. 5. meter of the circle ABC; and BC is also a diameter of it ; 4. therefore the point H where they meet is the centre of the circle, and consequently H is given : and the point D is given, wherefore DH is given in magnitude : and because AD is pa:

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