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If a straight line be drawn through a given point within a circle given in position, the rectangle con. tained 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

E 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 ; 8

A C • 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, contain: ing the given angle BAC, BC is given in magnitude: By. 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 iso 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 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,

E so is BA together with AC to BC;

В 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 con tained by BA and AC together, and DE, is given.

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Otherwise,

32. I.

Produce CA, and make AF equal to AB, and join BF; and because the angle BAC is double 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 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

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 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 G: And because BC cuts off from the circle ABC given in magnitude, the segment BAC containing a

F given angle, BC is therefore 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

E
ratio of BC to BD is given: And
because the angle CAE is equal
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:

2

B

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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 equiangular 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 triangle BCG 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, because its sides CB, BD are given : Therefore the rectangle contained by the excess of BA, AC, and the straight line DE is given

95.

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 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 iq the circumference, and let AE be drawn at right angles to 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 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 point Dis given, wherefore DH is given in magnitude. And because AD is parallel to FG, and GH equal to HA; DH is

0 4. 6. equal to HF, and AD equal to GF: And DH is given, therefore HF is given in magnitude; and it is also given A

E

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E

G * 80 Dat. in position, and the point H is given, therefore the point

F is given.

And because the straight line EFG is drawn from a given

point F without or within the circle ABC given in position, 95 or 96 therefored the rectangle EF, FG is given: And GF is Dat.

equal to AD, wherefore the rectangle AD, EF is given.)

PROP. C.

If from a given point in a straight line given in position, a straight line be drawn to any point in the circumference of a circle given in position; and from this point a straight line be drawn, making with the first an angle equal to the difference of a right angle, and the angle contained by the straight line given in position, and the straight line which joins the given point and the centre of the circle; and from the point in which the second line meets the circumference again, a third straight line be drawn, making with the second an angle equal to that which the first makes with the second: The point in which this third line meets the straight line given in position is given; as also the rectangle contained by the first straight line, and the segment of the third betwixt the circumference and the straight line given in position, is given.

Let the straight line CD be drawn from the giveni point C in the straight line AB given in position, to the circumference of the

circle DEF given in position, of which G is the centre; join CG, and from the point D let DF be drawn, making the angle CDF equal to the difference of a right angle, and the angle BCG, and from the point Flet FE be

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