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A

D

a. 46. I. b. 29. 6.

E

B

Upon AB defcribe the fquare BC, and to AC apply the paral- Book VI. lelogram CD equal to BC exceeding by the figure AD fimilar to in BC. but BC is a fquare, therefore alfo AD is a fquare. and because BC is equal to CD, by taking the common part CE from each, the remainder BF is equal to the remainder AD. and thefe figures are equiangular, therefore their fides about the equal angles are reciprocally proportional . wherefore as FE to ED, so AE to EB. but FE is equal to AC, that is to AB; and ED is equal to AE. there- C fore as BA to AE, fo is AE to EB. but AB is greater than AE; wherefore AE is greater than EB. therefore the straight line AB is cut in extreme and mean ratio in E f. Which was to be done.

Otherwife,

F

Let AB be the given ftraight line; it is required to cut it in extreme and mean ratio.

Divide AB in the point C, fo that the rectangle contained by AB, BC be equal to the fquare of AC 5. then

because the rectangle AB, BC is equal to the A fquare of AC, as BA to AC, fo is AC to

e. 14. 6. d. 34. 1.

c. 14. 5. f. 3. Def. 6.

8. 11. 2.

C B

CB b. therefore AB is cut in extreme and mean ratio in Cf. Which h. 17. 6. was to be done.

PROP. XXXI. THEOR.

IN right angled triangles the rectilineal figure describ- See M, ed upon the fide oppofite to the right angle, is equal to the fimilar, and fimilarly defcribed figures upon the fides containing the right angle.

Let ABC be a right angled triangle, having the right angle BAC. the rectilineal figure defcribed upon BC is equal to the fimilar and fimilarly defcribed figures upon BA, AC.

Draw the perpendicular AD; therefore becaufe in the right angled triangle ABC, AD is drawn from the right angle at A perpendicular to the bafe BC, the triangles ABD, ADC are fimilar to the whole triangle ABC, and to one another". and because the triangle a. . .

Book VI. ABC is fimilar to ABD, as CB to BA, fo is BA to BD b. and becaufe these three ftraight lines are proportionals, as the first to the third, fo is the figure upon the first to the fimilar, and fimilarly defcribed figure upon the fe

b. 4. 6.

c. 2. Cor. cond. therefore as CB to

20. 6.

d. B. 5.

e. 24 5.

f. A. 5.

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fo is the figure upon CA to that upon CB. Wherefore as BD and DC together to BC, fo are the figures upon BA, AC to that upon BC. but BD and DC together are equal to BC. Therefore the figure defcribed on BC is equal to the fimilar and fimilarly defcribed figures on BA, AC. Wherefore in right angled triangles, &c. Q. E. D.

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IF

PROP. XXXII. THEOR.

F two triangles which have two fides of the one proportional to two fides of the other, be joined at one angle fo as to have their homologous fides parallel to one another; the remaining fides fhall be in a ftraight line.

Let ABC, DCE be two triangles which have the two fides BA, AC proportional to the two CD, DE, viz. BA to AC, as CD to DE; and let AB be parallel to DC, and AC to DE. BC and CE are in a ftraight line.

Becaufe AB is parallel to DC, and the ftraight line. AC meets them, the alternate angles BAC, ACD are equal; for the fame reafon the angle CDE is equal to the angle ACD; wherefore alfo BAC is equal to CDE.

and because the triangles

B

C

E

ABC, DCE have one angle at A equal to one at D, and the fides about thefe angles proportionals, viz. BA to AC, as CD to DE,

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b

the triangle ABC is equiangular 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 b. 5. 6. 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 alfo c, 32. x. the angles ACE, ACB are equal to two right angles. and fince at the point C in the straight line AC, the two ftraight lines BC, CE, which are on the oppofite fides of it, make the adjacent angles ACE, ACB equal to two right angles; therefore 4 BC and CE are d. 14. 1. in a ftraight line. Wherefore if two triangles, &c. Q. E. D.

IN

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N equal circles, angles whether at the centers or cir- See N. cumferences have the fame ratio which the circumferences on which they fland have to one another. fo also have the fectors.

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, 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 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. there- a. 17. 3, fore what multiple foever the circumference BL is of the circumference BC, the fame multiple is the angle BGL of the angle BGC. for the fame reafon, whatever multiple the circumference EN is of

2. 27.3.

Book VI. the circumference EF, the fame multiple is the angle EHN of the angle EHF. and if the circumference BL be equal to the circumference EN, the angle BGL is alfo equal to the angle EHN; and if the circumference BL be greater than EN, likewise the angle BGL is greater than EHN; and if less, lefs. 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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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 lefs, lefs. as therefore the circumference BC b.s. Def. 5. to the circumference EF, fob is the angle BGC to the angle EHF. but as the angle BGC is to the angle EHF, fo is the angle BAC to the angle EDF, for each is double of each, therefore as the circumference BC is to EF, fo is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

C. 15. 5.

d. 20. 3.

C. 4.

Allo, as the circumference BC to EF, fo is the fector BGC to the fector EHF. Join BC, CK, and in the circumferences 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 that they contain equal angles; the base BC is equal to the bafe 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 a ; and f.1 Def 3. the fegment BXC is therefore fimilar to the fegment COK f; and they are upon equal ftraight lines BC, CK. but fimilar fegments of 3. circles upon equal ftraight lines, are equal to one another. there

fore the fegment BXC is equal to the fegment COK, and the tri- Book VI. angle 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 equalto each of the sectors BGC, CGK. in the fame manner the fectors EHF, FHM, MHN may be proved equal to one another. therefore what multiple foever the circumference BL is of the circumference BC, the fame multiple is the sector BGL of the fector BGC. for the fame reafon, whatever multiple the circumference EN is of EF, the fame multiple is the fector EHN of the fector EHF. and if the circumference BL be equal to EN, the fector BGL is equal to the fector EHN; and if the circumfe

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

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