Page images
PDF
EPUB
[ocr errors]
[ocr errors]

because the parallelogram GH is greater than
EL, therefore the side KH is greater than FL,
and KG than FE: produce FL and FE, and
make FLM equal to RH, and FEN to KG, and
complete the parallelogram MN: MN is there-
fore equal and similar to GH: but GH is simi-
lar to EL ; wherefore
MN is similar to EL; K

АН
and consequently EL
and MN are about the
same diameter:* draw

* 26.6. their diameter FX,

с

LM and complete the

E scheme. Therefore,

BIO since GH is equal to EL and C together,

N P X and that GH is equal to MN; MN is equal to EL and C: take away the common part EL; then the remainder, viz. the gnomon NOL, is equal to C. And because AE is equal to EB, the parallelogram AN is equal* to the parallelo- *36. 1. gram NB, that

to BM:* add NO to each ; *43. 1. therefore the whole, viz. the parallelogram AX, is equal to the gnomon NOL : but the gnomon NOL is equal to C; therefore also AX is equal to C. Wherefore to the straight line AB there is applied the parallelogram AX equal to the given rectilineal figure C, exceeding by the parallelogram PO, which is similar to D, because PO is similar* to EL. Which was to be done. *24.6.

[ocr errors]
[ocr errors]

PROP. XXX. PROB.
To cut a given straight line in extreme and mean

ratio.
Let AB be the given straight line; it is re-
quired to cut it in extreme and mean ratio.

Upon AB describe* the square BC, and to 46. 1. AC * apply the parallelogram CD, equal to BC, * 29. 6. exceeding by the figure AD similar to BC:

[ocr errors]

T

then, since BC is a square,

therefore also 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 equiangular, therefore their sides about the

D * 14.6 equal angles are reciprocally* pro

portional: therefore, as FE to

ED, so AE to EB: but FE is A * 31. 1. equal * to AC, that is, tot AB; +30 Def. and ED is equal to AE; therefore as BA to AE, so is AE to EB:

С F but AB is greater than AE ; *14.5. wherefore AE is greater than EB : therefore

the straight line AB is cut in extreme and mean **Def.6. ratio in E.* Which was to be done.

Otherwise, Let AB be the given straight line; it is required to cut it extreme and mean ratio.

Divide AB in the point C, so that the rectangle contained by AB, BC, may be equal

square

of AC: A CB then, because the rectangle AB, BC is equal to the square of AC; as BĂ to AC, so is AC to

CB : * therefore AB is cut in extreme and mean #3Def.6. ratio in C.* Which was to be done.

[ocr errors]

*

[ocr errors]

to the

* 17.6.

PROP. XXXI. THEOR.
In right angled triangles, the rectilineal figure

described upon the side opposite to the right
angle, is equal to the similar and similarly de-
scribed figures upon the sides containing the
right angle.

Let-ABC be a right angled triangle, having the right angle BAC: the rectilineal figure described upon BC shall be equal to the similar and similarly described figures upon BA, AC.

Draw the perpendicular + AD: therefore, because in the right angled triangle ABC, AD is

+12. 1.

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 *8.6 another: and because the triangle ABC is similar to ADB, as CB to BA, so is* BA to BD : *4.6 and because these three straight lines are proportionals, as the first is to the third, so is the figure upon the first to the similar and similarly described figure * upon the second: therefore as *2 Cor. CB to BD, so is the figure upon CB to the simi- 20.6. lar and similarly described figure upon BẢ: and inversely, * as DB to BC, so is the figure upon BA to that upon

BC: for the same Bf reason, as DC to CB, so is the figure upon CA to that upon CB : therefore as BD and DC together to

so are the figures upon BA, AČ to that *24. 5. upon BC: but BD and DC together are equal to BC; therefore the figure described on BC is equal to the similar and similarly described * A.5. figures on BA, AC. Wherefore, in right angled triangles, &c. Q.E.D.

[ocr errors]

* B. 5.

С

BC,

*

*

PROP. XXXII. THEOR.
If' tuo triangles which have two sides of the one

proportional to two sides of the other, be joined
at one angle so as to have their homologous
sides parallel to one another ; the remaining
sides shall 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

A CD to DE; and let AB be parallel to DC, and AC to DE: BC and CE shall be in a straight line.

Because AD is parallel to DC, and the straight line

*

*6.6

* 29. 1. AC meets them, the alternate angles * BAC,

ACD are equal; for the same reason, the angle

CDE is equal to the angle ACD; where+1 Ax. fore also BAC is equal t to CDE: 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 to AC, as CD to DĚ, the triangle ABC is equiangular * to DCE: 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 +2 Ax. is equal t to the two angles ABC, BAC: add the common

angle ACB, then the angles ACE, ACB are equal to the angles ABC, BĂC, ACB : but * 32. 1. ABC, BAC, ACB are equal * to two right angles;

therefore also 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 * BC and CE are in a straight line. Wherefore, if two triangles, &c Q. E. Ď.

:

:

14. 1.

PROP. XXXIII. THEOR.

In equal circles, angles, whether at the centres or

circumferences, have the same ratio which the circumferences on which they stand have to one another : 80 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 shall the angle BGC be 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 * 27. 3.
multiple soever 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
the circumference EF, the same multiple is the
angle EHN of the angle EHF: and if the cir-
cumference BL be equal to the circumference
EN, the angle BGL is also equal * to the angle * 27. 3.
EHN; and if the circumference BL be greater
than ÉN, likewise the angle BGL is greater
than EHN; and if less, less: therefore since
there are four magnitudes, the two circumfer-
ences BC, EF, and the two angles BGC, EHF;
and that of the circumference BC, and of the angle
BGC, have been taken any equimultiples what-
ever, viz. the circumference BL, and the angle
BGL; and of the circumference EF, and of the
angle EHF, any equimultiples whatever, viz.
the circumference EN, and the angle EHN;
and since it has been proved, that if the circum-
А

D
L

Н. N
K

M
B

E F

ference BL be greater than EN, the angle BGL is greater than EHN; and if equal, equal; and if less, less : therefore as the circumference BC to the circumference EF, so * is the angle BGC #5Def.5. to the angle EHF: but as the angle BGC is to the angle EHF, so is * the angle BAC to the *15.5. angle EDF: for each is double * of each ; there- * 20.3. fore, 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, so shall

*

« PreviousContinue »