given in magnitude cutting off the fegment BAC, containing the given angle BAC; BC is given in magnitude: By the a 91. dat. fame reafon BD is given; therefore the ratio of BC to BD b 1. dat. is given: And because the angle BAC is bifected by AD, as

с

b

e 21. 3.

BA to AC, fo is BE to EC; and, by permutation, as AB c 3. 6. to BE, fo is AC to CE; wherefored as BA and AC together d 12. 5. to BC, fo is AC to CE: And because the angle BAE is equal to EAC, and the angle ACE to F ADB, the triangle ACE is equiangu-" lar to the triangle ADB; therefore as AC to CE, fo is AD to DB: But as AC to CE, fo is BA together with AC to BC; as therefore BA and AC to BC, fo is AD to DB; and, by permutation, as BA and AC to AD, fo 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.

B

C

Alfo 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, fo is AC to CE; and as AC to CE, fo is BA and AC to BC; therefore as BA and AC to BC, fo 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.

Other wife.

a

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 angie BCA is equal to BDA, therefore the triangle FCB is equiangular to ABD: As therefore as FC to CB, fo is AD to DB; and, by permutation, as FC, that is, BA and AC together to AD, fo 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 EFC 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, fo is BD to DE; therefore the rectangle contained by FC, that is, BA and AC together, and DE is e

qual

P.

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

IF a ftraight line be drawn within a circle given in mag. nitude, cutting off a fegment containing a given angle; if the angle adjacent to the angle in the fegment be bifected by a straight line produced till it meet the circumference again and the base of the fegment; the excess of the straight lines which contain the given angle shall have a given ratio to the fegment of the bifecting line which is within the circle; and the rectangle contained by the fame excess and the fegment of the bifecting line betwixt the base produced and the point where it again meets the circumference, fhall be given.

Let the ftraight line BC be drawn within the circle ABC given in magnitude cutting off a fegment containing the given angle BAC, and let the angle CAF adjacent to BAC be bifected by the ftraight line DAE meeting the circumference again in D, and BC the bafe of the fegment produced in E; the excefs of BA, AC has a given ratio to AD; and the rectangle which is contained by the fame 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 feg

a

ment BAC containing a given ana 91. dat. gle, BC is therefore given in magnitude: By the fame reafon BD is given, because the angle BAD is equal to the given angle EAF; there-B fore the ratio of BC to BD is given: And because the angle CAE is equal to EAF, of which CAE is equal to

D

the alternate angle AGB, and EAF to the interior and oppofite angle ABG; therefore the angle AGB is equal to ABG, and the ftraight line AB equal to AG; so that GC is the excess

of

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 oppofite interior angle BDA of the quadrilateral BCAD in the circle; therefore the triangle BGC is equiangular to BDA: Therefore as GC to CB, fo is AD to DB; and, by permutation, as GC which is the excess of BA, AC to AD, fo is CB to BD: And the ratio of CB to BD is given; therefore the ratio of the excefs 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, fo is BD to DE; and confequently the rectangle GC, DE is equal to the rectangle CB, BD which is given, because its fides CB, BD are given: Therefore the rectangle contained by the excefs of BA, AC and the straight line DE is given.

PROP. XCIX.

IF from a given point in the diameter of a circle given in pofition, or in the diameter produced, a ftraight 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 pofition, or in BC produced, let the given point D be taken, and from D let a ftraight 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.

95.

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

rallel

b 4. 6.

rallel to FG, and GH equal to HA; DH is equal to HF, and AD equal to GF: And DH is given, therefore HF is given in

magnitude; and it is alfo given in pofition, and the point H is e 35. dat. given, therefore the point F is given.

C

And because the straight line EFG is drawn from a given point F without or within the circle ABC given in pofition, 495. or 96. therefore the rectangle EF, FG is given: And GF is equal to AD, wherefore the rectangle AD, EF is given.

dat.

Iti

PROP. C.

F from a given point in a straight line given in pofition, a straight line be drawn to any point in the circumference of a circle given in pofition; 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 ftraight line given in pofition, and the ftraight line which joins the given point and the centre of the circle; and from the point in which the fecond line meets the circumference again, a third straight line be drawn making with the fecond an angle equal to that which the first makes with the second: The point in which this third line meets the ftraight line given in pofition is given; as alfo the rectangle contained by the first straight Tine and the fegment of the third betwixt the circumfe rence and the straight line given in pofition, is given.

Let the ftraight line CD be drawn from the given point C in the ftraight line AB given in pofition, to the circumference of the circle DEF given in pofition, 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 F let FE be drawn making

the

the angle DFE equal to CDF, meeting AB in H: The point His given; as alfo the rectangle CD, FH.

Let CD, FH meet one another in the point K, from which draw KL D perpendicular to DF; and let DC meet the circumference again in M, and let FH meet the fame in E, and A join MG, GF, GH.

M

K

Because the angles MDF, DFE are equal to one another, the circumferences MF, DE are equal; and adding or taking away the common part ME, the circumference DM is equal to EF; therefore the ftraight line DM is equal to the ftraight line EF, and the angle GMD to the angle GFE ; and the angles GMC, GFH are equal E to one another, because they are either the fame with the angles GMD, GFE, or adjacent to them: And be cause the angles KDL, LKD are together equal to a right angle, that is, by the hypothefis, to the angles KDL, GCB; the angle GCB or GCH is & A C qual to the angle (LKD, that is, to

C

MY

D

F

HB

B

a 26.3.

b 8. 1.

C 32. I.

the angle) LKF or GKH: Therefore the points C, K, H, G are in the circumference of a circle; and the angle GCK is therefore equal to the angle GHF; and the angle GMC is equal to GFH, and the ftraight line GM to GF; therefore CG d 26. 1. is equal to GH, and CM to HF: And because CG is equal to GH, the angle GCH is equal to GHC; but the angle GCH is given: Therefore GHC is given, and confequently the angle CGH is given; and CG is given in pofition, and the point G ; therefore GH is given in pofition; and CB is alfo given in po- e 32. dat. fition, wherefore the point H is given.

And because HF is equal to CM, the rectangle DC, FH is e

qual to DC, CM: But DC, CM is given, because the point f 95. or 96. C is given; therefore the rectangle DC, FH is given.

FINIS.

dat.