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angle 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. 3

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Produce CA, and make AF equal to AB, and join BF; and because the angle BAC is double * of each of * 5. & 32.1. 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 AČB 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 mag

nitude, 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

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adjacent to BAC be bisected by the straight line DAE,
meeting the circumference again in D, and BC the
AC, has segment produced and
is contained by the
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

Secment
BAC containing a given an-
+91 Dat. gle, BC is therefore given *

in magnitude : by the same
reason BD is given, because
the angle BAD is equal to
the given angle EAF; there the
fore the ratio of BC to BD
is given : and because the
angle CAE is equal to EAF,
of which 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: and because the angle BGC is equal to GAE,
that is, to EAF, or the angle BAD; and that the an.
gle BCG is equal to the opposite interior angle BDĄ
of the quadrilateral BCAD in the circle; therefore
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 : suhe therefore the rectangle contained by the excess of BA,,

AC, and the straight line DE is given.

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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 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 diameter of the circle ABC; and BC is • Cor. 5. 4. 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 D is given, wherefore DH is given in magnitude. And because AD is parallel to FG, and GH equal to HA; DH is equal * to HF, 4. 6. and AD equal to GF: and DH is given, therefore HF is given in magnitude; and it is also given in po

A
E

CF
C
DB H

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BD

sition, and the point H is given, therefore * the point * 30. Dat. 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, therefore * the rectangle EF, FG is given : * 95 or 96 and GF is equal to AD, wherefore the rectangle AD, EF is given.

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3160 If from a given point in a straight line given in posiz

posi

to any L 91 point a straight line be drawn, making

with the first an angle equal to the difference of a right angle, and

to the poslo and the centre of the circle ; and from the point in

once 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 betrixt the straight line given in position, is given.

Let the straight line CD
be drawn from the given
point C, in the straight line
AB given in position, to the Devis pals at
circumference of the circle

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DEF given in position, of
which G is the centre; join

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CG, and from the point D
let DF be drawn making the

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E.
angle CDF equal to the dif-
ference of a right angle, and
the angle BCG, and from
the point Flet FE be drawn,
making the angle DFE, equal

D
to CDF, meeting AB in H:
The point H is given ; as
also the rectangle CD, FH.

E
Let CD, FH, meet one

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another in the point K, from

K
which draw KL perpendicu-
lar to DF; and let DC meet
the circumference again in MY
M, and let FH meet the

F
same in E, and join MG, AC

HB GF, GH.

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F

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Because the angles MDF, DFE, are equal to one another, the circumferences MF, DE, are equal *; and * 26. 3. adding or taking away the common part MĖ, the circumference DM is equal to EF; therefore the straight line DM is equal to the straight line EF, and the angle GMD to the angle * GFE; and the angles GMC, * 8. 1. eith equal to one another, because they are either the same with the angles GMD, GFE, or adjacent to them: and because the angles KDL, LKD, are together equal * to a right angle, that is, by the • 32. 1. hypothesis, to the angles KDL, GCB; the angle GCB or GCH is equal to the angle (LKD, that is to the angle) LKF or GHK: therefore the points C, K, H, G, are in the angle GCK is therefore equal to the angle GHF: and the angle GMC is equal to GFH, and the straight line GM to GF; therefore * CG is equal to GH, and CM * 26. 1. to HF: and because CG is equal to GH, the angle GCH is equal to GHC; but the angle GCH is given : therefore ĠHC is given, and consequently the angle CGH is given; and CG is given in position, and the point G; therefore * GH is given in position; and * 32 Dat. CB is also given in position, wherefore the point H is given.

And because HF is equal to CM, the rectangle DC, FH, is equal to DC, CM: but DC, CM is given *, because the point C is given, therefore the * 95 or 96 rectangle DC, FH is given.

Dat.

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