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e. 14. 6.

Upon AB describe the square BC, and to AC apply the paral- Book VI. lelogram CD equal to BC exceeding by the figure AD fimilar to m BC b. but BC is a square, therefore also

a. 46. I.

b. 29. 6.

D AD is a square. 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 these figures are A

B equiangular, therefore their fides about the equal angles are reciprocally proportional ". wherefore as FE to ED, fo AE to EB. but FE is equal to AC4, that is

d. 34. 1. to AB; and ED is equal to AE. there. C

F fore as BA to AE, fo is AE to EB. but AB is greater than AE; wherefore AE is greater than EB. there- c. 14. 5. fore the straight line AB is cut in extreme and mean ratio in E f, f. 3. Dcf. 6. Which was to be done.

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 5. then because the rectangle AB, BC is equal to the A СВ square of AC, as BA to AC, so is AC to CB h. therefore AB is cut in extreme and mean ratio in Ci. Which h. 17. 6, was to be done.

8. II. 2.

PRO P. XXXI. THEOR. IN N right angled triangles the rectilineal figure describ- See Y,

, ed upon the side 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, having the right angie BAC. the rectilineal figure described upon BC is equal to the similar and similarly described figures upon BA, AC.

Draw the perpendicular AD; therefore because in the right an. gled 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 another". and because the triangle a. . 6,

20. 6.

Book VI. ABC is similar to ABD, as CB to BA, so is BA to BD b. and beWcause these three straight lines are proportionals, as the first to the b. 4. 6. third, fo is the figure upon the first to the similar, and similarly de

fcribed figure upon the fec. 2. Cor. cond. therefore as CB to

BD, fo is the figure upon CB
to the similar and similarly

defcribed figure upon BA. d. B. s.

and, inversely, as DB to
BC, so is the figure upon B

D
BA to that upon BC. for the
same reason, 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. bat BD and DC together are equal to BC. Therefore the
figure defcribed on DC is equal to the fimilar and similarly describ-
ed figures on BA, AC. Wherefore in right angled triangles, &c.
0. E. D.

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1. A. 5.

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See N.

F
F 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 fides parallel to one another; the remaining fides ihall be in a straight line.

Let ABC, DCE be two triangles which have the two sides 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 straight line.
Because AB is parallel to

A
DC, and the straight line
AC meets them, the alter-
nate angles BAC, ACD are
equal; for the same reason
the angle CDE is equal to
the angle ACD; wherefore
also BAC is equal to CDE.

B

E and because the triangles ABC, DCE have one angle at A equal to one at D, and the sides about these angles proportionals, viz. BA - AC, as CD to DE,

A

1 29. I. :

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the triangle ABC is equiangular 5 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.6.6. 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 also c 32. I. the angles ACE, ACB are equal to two right angles. and since 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. Q. E. D.

PRO P. XXXIII. THEOR.

IN equal circles, angles whether at the centers or cir- Sce N.

cumferences have the same ratio which the circumferences on which they ftand have to one another. fo 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, fo is the angle BGC to the angle EHF, and the angle BAC to the angle EDF, and also the fector BGC to the sector EIF.

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 A

LD

H

N
K

M
B

E all equal, the angles BGC, CGK, KGL are also all equal". there- a. 27. 3, fore what multiple foever the circumference BL is of the circumference BC, the same multiple is the angle BGL of the angle BGC. for the same 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 nangle EHF. and if the circumference BL be equal to the circumfe.

rence EN, the angle BGL is also 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, 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 equimultiplcs whatever, viz. the circumference EN, and the angie EHN, and it has been proved that if the circumference BL be greater than EN, the angle BGL is greater than EAN; and if

equal, equal; and if less, leís. as therefore the circumference BC b.s. Def. s. 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 the angle BAC to
the angle EDF, for cach is double of each d. therefore as the cir-
cumference BC is to EF, fo is the angle BGC to the angle EHF,
and the angle BAC to the angle EDF.
i Alio, as the circumference BC to EF, so 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 iwo sides BG, GC are equal to the two
CG, GK, and that they contain equal angles; the base BC is equal o
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 a ;

and 1.11. Def. 3. the fegment BXC is therefore similar to the segment COK f; and

they are upon equal straight lines BC, CK, but similar segments of circles upon equal straight lines, are equal to one another, there.

c. 4. I.

24. 3

fore the segment 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 equalto each of the sectors BGC, CGK. in the same 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 same multiple is the sector BGL of the sector BGC. for the same reason, whatever multiple the circumference EN is of EF, the same multiple is the sector EHN of the sector EHF. and if the circumference BL be equal to EN, zhe sector BGL is equal to the sector EHN; and if the circumfe

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H. N
K
O

M
B
Х.

E F rence BL be greater than EN, the sector BGL is greater than the fector EHN; and if less, less. since 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 factor EHF, the circumference EN and factor EHN are any equimultiples whatever; and that it has been proved if the circumference BL be greater than EN, the fiétor BGL is greater than the sector EHN; and if equal, equal; and if less, less. Therefore b as the circumference BC is to the circumference EF, fo b. s. Def.s. is the sector BGC to the sector EHF. Wherefore in equal circles, &c. Q. E. D.

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