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Book VI. ved to be eqnal to ACD: Therefore the whole angle ACE is equal to the two angles ABC, BAC; add the common angle ACB, then the angles ACE, ACB are equal to the angles ABC, BAC, ACB: But ABC, BAC, ACB are equal 32. I. to two right angles c; therefore alfo the angles ACE, ACB are equal to two right angles: And fince at the point C, in the ftraight line AC, the two ftraight lines BC, CE, which are on the oppofite fides of it, make the adjacent angles d 14. I. ACE, ACB equal to two right angles; therefore d BC and CE are in a ftraight line. Wherefore, if two triangles, &c. Q. E. D.

PROP. XXXIII. THEOR.

N equal circles, angles, whether at the centres or circumferences, have the fame ratio which the arches, on which they ftand, have to one another: So also have the fectors.

Let ABC, DEF be equal circles; and at their centres the angles BGC, EHF, and the angles BAC, EDF at their circumferences; as the arch BC to the arch EF, fo is the angle BGC to the angle EHF, and the angle BAC to the angle EDF; and alfo the fector BGC to the fector EHF.

Take any number of arches CK, KL, each equal to BC, and any number whatever FM, MN each equal to EF; and join GK, GL, HM, HN. Because the arches BC, CK, KL are all equal, the angles BGC, CGK, KGL are also all e27. 3. qual a: Therefore, what multiple foever the arch BL is of the arch BC, the fame multiple is the angle BGL of the angle BGC: For the fame reason, whatever multiple the arch EN is of the arch EF, the fame multiple is the angle EHN of the angle EHF. But if the arch BL be equal to the arch EN, the angle BGL is alfo equal a to the angle EHN; or if the arch BL be greater than EN, likewife the angle BGL is greater than EHN: and if lefs, lefs: There being then four magnitudes, the two arches BC, EF,

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and the two angles BGC, EHF, and of the arch BC, and of Book VL `the angle BGC, have been taken any equimaltiples whatever, viz. the arch BL, and the angle BGL; and of the arch EF, and of the angle EHF, any equimultiples whatever, viz.

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the arch EN, and the angle EHN: And it has been proved, that, if the arch BL be greater than EN, the angle BGL is greater than EHN; and if equal, equal; and if lefs, lefs: As therefore, the arch BC to the arch EF, fob is the angle b5. def. 5. BGC to the angle EHF; But as the angle BGC is to the angle EHF, fo isc the angle BAC to the angle EDF, for c 15. 5. each is double of eachd: Therefore, as the circumference d 20. 3. BC is to EF, fo is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

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Alfo, as the arch BC to EF, fo is the sector BGC to the fector EHF. Join BC, CK, and in the arches BC, CK take any points X, O, and join BX, XC, CO, OK: Then, because in the triangles GBC, GCK, the two fides BG, GC are equal to the two CG, GK, and alfo contain equal angles; the base BC is equal to the base CK, and the tri- e 4. 1. angle GBC to the triangle GCK: And because the arch BC is equal to the arch CK, the remaining part of the whole circumference of the circle ABC, is equal to the remaining part of the whole circumference of the fame circle : Wherefore the angle BXC is equal to the angle COK; and the fegment BXC is therefore fimilar to the fegment COK f; and they are upon equal ftraight lines BC, CK: But fimilar segments of circles upon equal ftraight lines, are equal g to one another: Therefore the fegment BXC is equal g 24. 3.

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Book VI. to the fegment COK: And the triangle BGC is equal to the triangle CGK; therefore the whole, the fector BGC is equal to the whole, the sector CGK. For the fame reason, the sector KGL is equal to each of the sectors BGC, CGK; and in the same manner, the sectors EHF, FHM, MHN may be proved equal to one another: Therefore, what multiple foever the arch BL is of the arch BC, the fame multiple is the fector BGL of the fector BGC. For the fame reafon, whatever multiple the arch EN is of EF, the fame multiple is the fector EHN

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of the sector EHF: Now if the arch BL be equal to EN, the fector BGL is equal to the fector EHN; and if the arch BL be greater than EN, the fector BGL is greater than the fector EHN; and if lefs, lefs: Since, then, there are four magnitudes, the two arches BC, EF, and the two fectors BGC, EHF, and of the arch BC, and fector BGC, the arch BL and the sector BGL are any equimultiples whatever; and of the arch EF, and fector EHF, the arch EN and fector EHN, are any equimultiples whatever; and it has been proved, that if the arch BL be greater than EN, the fector BGL is greater than the fector EHN; if equal, equal; and if lefs, lefs; b 5. def. 5. therefore b, as the arch BC is to the arch EF, fo is the fector BGC to the fector FHF. Wherefore, in equal circles, &c. Q. E. D.

PROP.

Book VI.

IF

PROP. B. THEOR.

F an angle of a triangle be bifected by a straight line, which likewife cuts the bafe; the rectangle contained by the fides of the triangle is equal to the rectangle contained by the fegments of the base, together with the fquare of the ftraight line bifecting the angle.

Let ABC be a triangle, and let the angle BAC be bifected by the straight line AD; the rectangle BA.AC is equal to the rectangle BD.DC, together with the square of AD.

B

/D

a 5.4.

b 21.3.

Defcribe the circle ACB about the triangle, and produce AD to the circumference in E, and join EC. Then, because the angle BAD is equal to the angle CAE, and the angle ABD to the angle AEC, for they are in the fame fegment; the triangles ABD, AEC are equiangular to one another: Therefore BA: AD :: EA:c AC, and confequently, BA.AC & AD.AE-ED.DA e +DA2. But ED.DA➡ d 16. 6. BD.DC, therefore BA.ACBD.DC+DA2. Wherefore, if e 3. 2. an angle, &c. Q. E. D.

E

c 4. 6.

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Book VI.

PROP. C.

THEOR.

I rectangle

from any angle of a triangle a straight line be

5.4.

b 3.3.

contained by the fides of the triangle is equal to the rectangle contained by the perpendicular, and the diameter of the circle defcribed about the triangle.

Let ABC be a triangle, and AD the perpendicular from the angle A to the bafe BC; the rectangle BA.AC is equal to the rectangle contained by AD and the diameter of the circle defcribed about the triangle.

Defcribe the circle ACB about the triangle, and draw its diameter AE, and join EC: Becaufe the right angle BDA is equal to the angle ECA in a femicircle, and the angle ABD to the angle AEC, in the fame-fegc 21. 3. ment c; the triangles ABĎ, AEC are equiangular: Therefore, as BA to AD, fo is EA to AC: and confequently the

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d 4. 6.

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€ 16. 6. rectangle BA.AC is equal to

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the rectangle EA.AD. If, therefore, from an angle, &c. Q.E. D.

See N.

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PROP. D. PROB.

THE rectangle contained by the diagonals of a quadrilateral infcribed in a circle, is equal to both the rectangles contained by its oppofite fides.

Let ABCD be any quadrilateral infcribed in a circle,

and

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