A

189

Upon AB defcribe a the fquare BC, and to AC apply the Book VI. parallelogram CD equal to BC, exceeding by the figure AD fin milar to BC b: But BC is a fquare, therefore alfo AD is a fquare; and becaufe BC is equal to CD, by taking the common part CE from each, the remainder BF is equal to the remainder AD: And these figures are equiangular, therefore their fides about the equal angles are reciprocally proportional e : Wherefore, as FE to ED, fo AE to EB: But FE is equal to AC d, that is, to AB; and ED is equal to AE: Therefore as BA to AE, fo is AE to EB: But AB is greater than AE; wherefore AE is

a 46. I. b 29. 6.

D

E B

c 14. 6.

d 34. I.

F

greater than EB : Therefore the straight line AB is cut in ex- e 14. 5. treme and mean ratio in E f. Which was to be done.

Otherwife,

Let AB be the given straight 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 8.

f 3. def. 6.

g II. 2.

Then, because the rectangle AB, BC is equal A

B

to the fquare of AC, as BA to AC, fo is

AC to CBh: Therefore AB is cut in extreme and mean ra- h 17. 6. tio in Cf. Which was to be done.

IN

N right angled triangles, the rectilineal figure defcri- See N. bed upon the fide oppofite to the right angle, is equal

to the fimilar, and fimilarly defcribed figures upon

fides containing the right angle.

the

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

Draw the perpendicular AĎ; therefore, because 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 3, a 8. 6.

and

C

Book VI. and because the triangle ABC is fimilar to ADB, as CB to BA, fo is BA to BD; and because these three ftraight lines are proportionals, as the firft to the third, fo is the figure upon the first to the fimilar, and fimilarly defcribed figure upon the fecond: Therefore as CB to

b 4. 6.

c 2 Cor. 20. 6.

B. 5.

€ 24. 5. f A. 5.

BD, fo is the figure upon
CB to the fimilar and fimi-
larly defcribed figure upon
BA: And, inversely, as DB
to BC, fo is the figure upon
BA to that upon BC: For
the fame reafon, as DC to
CB, 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 f to the fimilar and fimilarly defcribed figures on BA, AC. Wherefore, in right angled triangles, &c. Q. E. D.

See-N.

a 29. I.

IF

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 straight 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.

Because AB is parallel to
DC, and the ftraight line A
AC meets them, the al-
ternate angles BAC, ACD
are equal; for the fame
reafon, the angle CDE is
equal to the angle ACD;
wherefore alfo BAC is e-
qual to CDE: And becaufe

191

the triangles ABC, DCE have one angle at A equal to one at Book VI. D, and the fides about these angles proportionals, viz. BA to AC, as CD to DE, the triangle ABC is equiangular to DCE: b 6. 6. Therefore the angle 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 alfo the angles ACE, c 32,5.. 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 ACE, ACB equal to two right angles; therefore 4 BC and d 14. 1. 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 cir- See N cumferences, have the fame ratio which the circumferences 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 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 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 all equal, the angles BGC, CGK, KGL are alfo all equal: Therefore what multiple foever the circum- a 27. 3. ference BL is of the circumference BC, the fame multiple is the angle BGL of the angle BGC: For the fame reason, whatever multiple the circumference EN is of the circumference EF, the fame multiple is the angle EHN of the angle EHF:

And

a 27. 3.

Book VI. 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 lefs, 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 circum. ference EF, and of the angle EHF, any equimultiples what

ever, 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 b5. Def. 5. BC 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 e 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. 6. d 20. 3.

c'4. I.

с

Alfo, as the circumference BC to EF, fo is the sector 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 bafe 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; f11. def. 3, and the fegment BXC is therefore fimilar to the fegment COKƒ;

and

and they are upon equal ftraight lines BC, BK: But fimilar feg. Book VI. ments of circles upon equal ftraight lines, are equal to one another: Therefore the fegment BXC is equal to the fegment COK: 8 24. 3. 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 fectors BGC, CGK: In the fame manner, the sectors 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 fector 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 circumference BL be greater than EN, the sector BGL is greater than the sector EHN; and if lefs, lefs: Since, then, there are four magnitudes, the two circumferences BC, EF, and the two feccors 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 fector 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, &c. Q. E. D.