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Upon AB describe a the square BC, and to AC apply the Book VI. parallelogram CD equal to BC, exceeding by the figure AD fie imilar to BC b: But BC is a square,
a 46. 1. therefore also AD is a fquare ; and be
b 29. 6.
D cause 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 equiangu
A lar, therefore their sides about the equal angles are reciprocally proportional c:
C 14. 6. Wherefore, as FE to ED, fo AE to EB : But FE is equal to AC 4, that is, to AB;
d 34. 1. and ED is equal to AE : Therefore as BA to AE, so is AE to EB : But AB is
F greater than AE; wherefore AE is greater than EB °: Therefore the straight line AB is cut in ex• e 14. 5. treme and mean ratio in El. Which was to be done.
f 3. def, 6. Otherwise, Let AB be the given straight line; it is required to cut it in extreme and mean ratio.
Divide AB in the point C, so that the rectangle contained by AB, BC be equal to the square of AC 8.
& II. 26 Then, because the rectangle AB, BC is equal A C B to the square 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 Ci. Which was to be done.
IN cight angled triangles, the rectilineal figure descri. See N.
bed upon the fide opposite to the right angle, is equal to the similar, and similarly described figures upon the sides containing the right angle.
Let ABC be a right angled triangle, baving the right angle BAC: The rectilineal figure described upon BC is equal to the limilar, and Gmilarly described 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 base BC, the triangles ABD, ADC are similar to the whole triangle ABC, and to one anocher, 1 8. 6.
Book VI. and because the triangle ABC is similar to ADB, as CB tá
BA, fo is BA to BD 6; and because these three straight lines b 4. 6.
are proportionals, as the first to the third, so is the figure up
on the first to the similar, and similarly described figure upon the ci Cor. second c: Therefore as CB to
BD, so is the figure upon
as BD and DC together to BC, so are the figures upon BA, AC € 24. 5. to that upon BC: But BD and DC together are equal to BC.
Therefore the figure described on BC is equalf to the similar and aimilarly described figures on BA, AC. Wherefore, in right angled triangles, &c. Q. E. D.
two triangles which have two sides of the one pro
portional to two sides of the other, be joined at one angle, so as to have their homologous sides parallel to one another ; the remaining fides shall be in a straight line.
Let ABC, DCE be two triangles which have the two sides
Because AB is parallel to
& 29. 1.
the triangles ABC, DCE have one angle at A equal to one at Book VL. D, and the fides about these angles proportionals, viz. BA to AC, as CD to DE, the triangle ABC is equiangular Oto 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 also the angles ACE, C 32. I. ACB are equal to two right angles : And lince 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 a BC and d 14. I. CE are in a straight line. Wherefore, if two triangles, &c. Q. E. D.
cumferences, have the same ratio which the circumferences on which they ítand have to one another : So also have the sectors.
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, 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 all equal, the angles BGC, CGK, KGL are also all equal : Therefore what multiple foever the circum- a 27. 3. ference BL is of the circumference BC, the same multiple is the angle BGL of the angle BGC: For the same reason, whatcver multiple the circumference EN is of the circumference EF, the fame multiple is the angle EHN of the angle EHF:
Book VI. And if the circumference BL be equal to the circumference
EN, the angle BGL is also equal to the angle EHN; and a 27. 3. 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 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 é.
qual, equal; and if less, lefs : As therefore the circumference bs. Def. 5. BC to the circumference EF, so b is the angle BGC to the
angle EHF: But as the angle BGC is to the angle EHF, so is c 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.
Also, as the circumference BC to EF, fo is the sector BGC to the sector 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 e6'4. I. qual 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; f11. def. 3. and the segment BXC is therefore similar to the segment COK,
and they are upon equal straight lines BC, BK: But Gmilar seg. Book vi.
fector BGL is equal to the sector EHN; md if the circumfe.