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b. 6. 6.

the triangle ABC is equiangularb to DCE. therefore the angle Book VI. ABC is equal to the angle DCE. and the angle BAC was proved to be equal 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 to two right angles; therefore also c. 32. 1. the angles ACE, ACB are equal to two right angles. and fince at the point C in the straight line AC, the two straight lines BC, CE, which are on the opposite sides of it, make the adjacent angles ACE, ACB equal to two right angles; therefore d BC and CE are d. 14. 1.

in a straight line. Wherefore if two triangles, &c.

IN

PROP. XXXIII. THEOR.

E. D.

N equal circles, angles whether at the centers or cir. See N. cumferences have the fame ratio which the circumferences on which they stand have to one another. so also have the sectors.

Let ABC, DEF be equal circles; and at their centers the angles BGC, EHF, and the angles BAC, EDF at their circumferences. as the circumference BC to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF; and also the sector BGC to the sector EHF.

Take any number of circumferences CK, KL each equal to BC, and any number whatever FM, MN each equal to EF; and join GK, GL, HM, HN. Because the circumferences BC, CK, KL are

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all equal, the angles BGC, CGK, KGL are alfo all equal a. there- a. 27. 3+

fore what multiple foever the circumference BL is of the circumfe

rence BC, the fame multiple is the angle BGL of the angle BGC.

for the fame reason, whatever multiple the circumference EN is of

2. 27.3.

Book VI. the circumference EF, the same multiple is the angle EHN of the angle FHF. and if the circumference BL be equal to the circumference EN, the angle BGL is alfo equal a to the angle EHN; and if the circumference BL be greater than EN, likewise the angle BGL is greater than EHN; and if less, less. there being then four magnitudes, the two circumferences BC, EF, and the two angles BGC, EHF; of the circumference BC, and of the angle BGC, have been taken any equimultiples whatever, viz. the circumference BL, and

the angle BGL; and of the circumference EF, and of the angle

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C. 15.5.

d. 20. 3.

EHF, any equimultiples whatever, viz. the circumference EN, and the angle EHN. and it has been proved that if the circumference BL be greater than EN, the angle BGL is greater than EHN; and if equal, equal; and if less, less. as therefore the circumference BC

b

b. 5. Def. 5. to the circumference EF, so is the angle BGC to the angle EHF. but as the angle BGC is to the angle EHF, so is the angle BAC to the angle EDF, for each is double of each d. therefore as the circumference BC is to EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

e. 4. 1.

Also, as the circumference BC to EF, so is the sector BGC to the sector EHF. Join BC, CK, and in the circumferences BC, СК 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 that they contain equal angles; the base BC is equal to the base CK, and the triangle GBC to the triangle GCK. and because the circumference BC is equal to the circumference 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 f.11.Def. 3. the segment BXC is therefore similar to the segment COK ; and they are upon equal straight lines BC, CK. but fimilar segments of 8.24. 3. circles upon equal straight lines, are equal & to one another. there. fore the fegment BXC is equal to the segment COK. and the tri- Book VI. angle BGC is equal to the triangle CGK; therefore the whole, the sector BGC is equal to the whole, the sector CGK. for the same reason the sector KGL is equal to each of the sectors BGC, CGK. in the fame manner the sectors EHF, FHM, MHN may be proved equal to one another. therefore what multiple soever the circumference BL is of the circumference BC, the fame multiple is the sector BGL of the sector BGC. for the fame reason, whatever multiple the circumference EN is of EF, the same multiple is the sectorEHN of the sector EHF. and if the circumference BL be equal to EN, the sector BGL is equal to the sector EHN; and if the circumfe

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rence BL be greater than EN, the sector BGL is greater than the sector EHN; and if less, less. fince then there are four magnitudes, the two circumferences BC, EF, and the two sectors BGC, EHF, and of the circumference BC and sector BGC, the circumference BL and sector BGL are any equimultiples whatever; and of the circumference EF and sector EHF, the circumference EN and sector EHN are any equimultiples whatever; and that it has been proved if the circumference BL be greater than EN, the sector BGL is greater than the sector EHN; and if equal, equal; and if less, less. Therefore as the circumference BC is to the circumference EF, so b. 5. Def. 5. is the sector BGC to the sector EHF. Wherefore in equal circles, &c. Q. E. D.

Book VL

See N.

I

PROP. B. THEOR.

F an angle of a triangle be bisected by a straight line,

which likewise cuts the base; the rectangle contained by the fides of the triangle is equal to the rectangle contained by the segments of the base, together with the square of the straight line bisecting the angle.

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

2.5.4.

с. 4. 6.

d. 16. 6.

e. 3. 2.

Describe the circle a 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 b. 21. 3. angle ABD to the angle b AEC, for they are in the same segment; the triangles ABD, AEC are equiangular to one another, therefore as B BA to AD, so is EA to AC, and confequently the rectangle BA, AC is equal d to the rectangle EA, AD, that is to the rectangle ED, DA together with the square of

A

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C

D

E

f. 35.3.

See N.

AD. but the rectangle ED, DA is equal to the rectangle & BD,
DC. Therefore the rectangle BA, AC is equal to the rectangle BD,
DC together with the square of AD. Wherefore if an angle, &c.

I

E. D.

F from

PROP. C. THEOR.

an angle of a triangle a straight line be drawn perpendicular to the base; the rectangle contained by the fides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle.

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

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equal to the rectangle EA, AD. If therefore from an angle, &c. c. 16. 6. Q. E. D.

T

1

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a qua

HE rectangle contained by the diagonals of
drilateral infcribed in a circle, is equal to both

the rectangles contained by its opposite sides.

Let ABCD be any quadrilateral inscribed in a circle, and join AC, BD; the rectangle contained by AC, BD is equal to the two rectangles contained by AB, CD and by AD, BC *.

4

Make the angle ABE equal to the angle DBC; add to each of these the common angle EBD, then the angle ABD is equal to the angle EBC. and the angle BDA is equal to the angle BCE, because a. 21. 3. they are in the same segment; therefore the triangle ABD is equi

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BA, DC is equal to the rectangle BD, AE. but the rectangle BC, AD has been shewn equal to the rectangle BD, CE; therefore the whole rectangle AC, BD is equal to the rectangle AB, DC together with the rectangle AD, BC. Therefore the rectangle, &c. Q. E. D.

* This is a Lemana of Cl. Ptolemaeus in page 9. of his μεγάλη σύνταξις.

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