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

d 28 Dat.

· PROP. XCIV.

91. 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 givena in position and magni. " B_ :""29 Dat. tude: And DBA is a rightb angle,

b 18. 3. wherefore DA is a diameter of the

"Cor. 5.4. circle DBA, described about the triangle DBA; and that circle is therefore given in position : And the circle BC is givend in position, there

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

" PROP. xcv. 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 position, cutting it in B, C; the rectangle BA, AC, is given. From the point A drawa 'AD

* 17. 3. touching the circle; therefore ADC is given in position and magnitude: And because AD is given, the square of AD is givenc, which is

56 Dat. equald to the rectangle BA, AC: Therefore the rectangle a BA, AC, is given.

94.

b 94 Dat.

36. 3.

: 93. . .

PROP. XCVI. m esto, 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, and produce it to the points E, F:

Because the points A, D are given, the a 29 Dat. straight line AD is given a in position ;

D/ and the circle BEC is given in position; si 28 Dat, therefore the points E, F, are givenb;

and the point A is given, therefore EA,

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.

94.

PROP. XCVIL "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

given in magnitude cutting off the segment BAC, contains ing the given angle BAC; BC is given a in magnitude: By • 91 Dar. 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 an-
gle ACE to ADBę, the triangle

21. 3. 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, 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 cona tained by BA and AC together, and DE, is given.

Otherwise,

Produce CA, and make AF equal to AB, and join BF; and because the angle BAC is double a of each of the angles. 3 5. BFA, BĄD; the angle BFA is equal to BAD; and the

*{32. T. 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 to. gether, 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

DE is equal to the rectangle CB, BD, which is given, and

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

P. : : . PROP. XCVIII.

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 bisecting 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 DAE, 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 Ğ: And because BC cuts off from the circle ABC given in magnitude, ID 1940 ai 10 the segment BAC containing a

given angle, BC is therefore 2 91 Dat. given a in magnitude: By the

same reason BD is given, because
the angle BAD is equal to the
given angle EAF; therefore the BS

uborg E.
ratio of BC to BD is given: And
because the angle CAE is equal DUNG persib
to EAF, of wbich CAE is equal to the alternate angle AGB,
and EAF to the interior and opposite angle ABG ; there-
fore the angle AGB is equal to ABG, and the straight line
AB equal to AG; so that GC is the excess of BA, AC:

excess and the rectancess of BA, PASE

And because the angle BGC is equal to GAE, that is, to
EAF, or the angle BĂD; 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 equi-
angular to BDA. 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 BC 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 tri
angle RCG is equiangular to 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, be-
cause its sides CB, BD are given : Therefore the rectangle
contained by the excess of BA, AC, and the straight line
DE is given.

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.: PROP. XCIX.
IF from the 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 circum.
ference, 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 iq
the circumference, and let AE be drawn at right angles to
DA, and from the point E where it meets the circumfer-
ence 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 isa the Cor. 5. 4. diameter of the circle ABC; and BC is also a diameter of it; therefore the point H, where they meet, is the centre of the circle, and consequently H is given : And the poim D-is given, wherefore DH is given in magnitude. And because AD is parallel to FG, and GH equal to HA; DH is

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