Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

T

PROPOSITION I. PROBLEM.

defcribe an equilateral triangle upon a given right
line.

Let AB be the given right line, upon which it is required to defcribe an equilateral triangle.

Book I.

About the center A, with the distance AB, describe the circle BCD2; and about the center B, with the distance BA, describe a Postulate the circle ACEb; from the point C, where the two circles interfect each other, draw the lines CA, CB".

3.

b Poft. I.

Then, because A is the center of the circle DBC, AC is equal to AB ; and because B is the center of the circle ACE; BC c Definitiis equal to BA, but CA is proved equal to AB, and BC to AB; on 15. therefore BC is equal to AC d: Therefore the three fides AB, dAxiom 1. BC, CA, are equal to one another: Therefore, upon the given right line AB, there is described an equilateral triangle ABC: Which was required.

A

PRO P. II. PRO B.

Ta given point to put a right line equal to a given
right line.

Let A be the given point, and BC the given right line, it is required to put a right line at the point A, equal to the given right line BC.

C I.

With the center C and distance BC, describe the circle BGH'; a Poft. 3. join AC, upon which defcribe an equilateral triangle DAC,b Poft, i. produce DC that paffes through the center to G, in the circumference; and DA to any distance Ed; with the center D, and d Post. z. diftance DG, defcribe the circle KGL

Then, because C is the center of the circle BGH, BC is equal

to CG; and because D is the center of the circle KGL, DG e Def. 15. is equal to DL, but DC is equal to DA: Therefore the remainders AL, and CG, are equalf; but BC, AL, are each f Ax. 3. proved equal to CG; and therefore equal to one another: 8 Ax. 1. Therefore, from the point A, there is drawn the right line AL, equal to the given right line BC: Which was required.

PROP.

Book I.

a 2.

b Poft. 3.

c Def. 15.

d Ax. I.

PRO P. III. PRO B.

WO unequal right lines being given, to cut off from the

Tgreater a part equal to the lejer.

Required to cut off from the greater AB a part AE, equal to the leffer C.

From the point A draw a right line AD equal to Ca, about the center A, with the distance AD, describe a circle DEFь; then, because A is the center of the circle DEF, AE is equal to AD, but AD is equal to C: Therefore AE is likewife equal to Cd: Which was required.

I

PRO P. IV. THEOREM.

F there be two triangles having two fides of the one equal to two fides of the other, each to each; and the angle contained by the two fides of the one, equal to the angle contained by the correfpondent fides of the other; then the base of the one triangle will be equal to the base of the other; the two triangles will be equal, and the remaining angles of the one equal to the remaining angles of the other, each to each, which the equal fides fubtend

Let the two triangles be ABC, DEF, having the two fides AB, AC, equal to the two fides DE, DF, each to each; that is, AB equal to DE, and AC equal to DF, and the angle BAC equal to EDF; then the bafes BC and EF will be equal, the triangle ABC equal to the triangle DEF, the angle ABC equal to the angle DEF; and ACB equal to DFE, each to each, which the equal fides fubtend.

For, let the triangle ABC be applied to the triangle DEF, fo as the point A may coincide with the point D, the right line AB with DE, then the point B will coincide with the point E; for AB and DE are equal; and, because the angles BÁC EDF are equal, the right line AC will coincide with DF, and the point C with F; for AC and DF are equal: Then, because the point B coincides with the point E, and C with F, the right lines BC and EF, will coincide"; and therefore equal: Thereb Ax. 8. fore the triangles ABC, DEF, are equal, the angle ABC equal to the angle DEF, and ACB to DFE, each to each, which the equal fides fubtend. Wherefore, &c.

a Def. 4.

T

PRO P. V. THE O R.

HE angles above the base of every ifofceles triangle are e-
qual to one another.

Let the triangle ABC be the ifofceles triangle, having the fide AB equal to the fide AC, then the angle ABC will be equal to the angle ACB.

Book I.

For, let a triangle DEF be likewife given, having the fides DE, DF, equal to the fides AB, AC, each to each; and the angle BAC equal to the angle EDF, then the bafes EF, BC, are equal; and the angle ABC equal to the angle DEF, and ACB to DFE, each to each; which the equal fides fubtend; but the fides AB, DF, are equal; therefore the angles a 4. ACB, DEF, are likewife equal; but the angles DEF, ABC, are b Ax. 1. equal; therefore the angles ABC, ACB, are likewife equal. Wherefore, &c.

CORROLLARY. If any triangle is equilateral, it will likewife be equiangular.

PRO P. VI. THE O R.

I fides fubtending these angles will like wife be equal

F any triangle have two angles in it equal to one another, the

Let the triangle ABC have the angles ABC, ACB, equal; then the fides AB, AC, will likewife be equal to one another.

For, if AB is not equal to AC, let one of them, as AB, be the greater; from which cut off DB equal to AC, and join DC; a 3. then, because DB is equal to AC, the two fides DB, BC, are equal to the two fides AC, BC, and the angles ACB, DBC, equal; therefore the bafe AB is equal to the base DC, and the triangles ABC, DBC, equal; that is, a part equal to the whole; b 4. which is abfurd: Wherefore the fides AB, AC, are equal; that is, the triangle is ifofceles. Wherefore, &c.

COR. Hence every equiangular triangle is alfo equilateral.

[ocr errors]

PRO P. VII. THE OR,

F from the extremity of any right line, to two different points on the fame fide, there be drawn two right lines equal to one another, the lines drawn from the other extremity, to the fame points, cannot be equal to one another.

If from the extremity A, of the right line AB, to the points C,D,. on the fame fide, the two lines AD, AC, are drawn equal

to

Book. I. to one another, from the extremity B, to the fame points C, D, the lines BC, BD, are not equal to one another.

a 5.

Join DC; for, because AC, AD, are equal, the angles ADC, ACD are equal; but the angle BDC is greater than ADC, or b Ax. 9. ACD, and much greater than BCD, but, if BD is equal to BC, the angles BCD, BDC, are equal, and likewise greater; which is impoffible: Therefore, BD is not equal to BC. If BD is made equal to BC, it is proved in the fame manner that AC is not equal to AD. Wherefore, &c.

[merged small][merged small][merged small][ocr errors]

I

PRO P. VIII. THE OR.

F two triangles have two fides of the one, equal to two fides of the other, and the base of the one equal to the base of the other, then the angles that these equal bafes fubtend will be equal to one

another.

Let the two triangles be ABC, DEF, having the two fides AB, AC, equal to the two fides DE, DF, each to each; and the bafes BC, EF, equal; then the angles BAC, ELF, will be equal.

For, let the triangle ABC be applied to the triangle DEF, fo that the right line BC may coincide with EF; then the point B will coincide with the point E, and C with F; for the fides BC, EF, are equal: And the fides BA, AC, will coincide with ED, DF, and the point A with D. If not, let the point A fall in G; then EG, ED, are equal; for each are equal to BA "; and FG, FD, will likewise be equal, which is impoffible. Wherefore the point A cannot fall in G, and, for the fame reason, in no point but D; therefore the angle BAC is equal to the angle EDF. Wherefore, &c.

To cut a

PRO P. IX. PRO B.

O cut a given right lined angle into two equal parts.

Let BAC be the given right lined angle required to be cut into two equal parts.

Affume any point D, in the right line AB, and cut off AE equal to AD; join DE, upon which defcribe the equilateral triangle DEF, and join AF; then the right line AF will bifect the given angle BAC; for, becaufe DA is equal to EA, and AF is common, and the bafe DF equal to EF; therefore the angle DAF is equal to the angle EAF: Therefore the angle BAC is bifected by the right line AF. Which was required.

PROP. X. PRO B.

BOOK I.

To cut

O cut a given finite right line into two equal parts.

Let AB be a given right line, required to be cut into two equal parts; upon it defcribe an equilateral triangle ABC; bifect the angle ACB by the right line CD; then is the right a

line AB bifected in D.

For, because AC is equal to CB, and CD common, and the angles ACD, BCD, equal, the bafes AD, BD, are equal b: b Therefore the right line AB is bifected in D. Which was re-. quired.

PRO P. XI. PRO B.

O draw a line at right angles to a given right line from à

T paint given in the Jame.

Let. AB be the given right line, and C the given point in it, from which it is required to draw a right line, at right angles to the given right line AB.

a

9.

Affume any point D in AC, and make CE equal to CDa; 2 3. upon DE defcribe an equilateral triangle DEF", and join FC, b 1. which will be at right angles to AB. For, because DC, CE, are equal, and CF common, the two fides DC, CF, are equal c. conit, to the two fides EC, CF, and the bafes FD, FE, equal, the angles FCD, FCE, are likewife equal: Therefore each of them & 8. is a right angle, and FC perpendicular to AB. Which was Def. 10 required.

T

PRO P. XII. PRO B.

O draw a right line perpendicular to a given indefinite right
line from a given point out of it:

Let AB be the given right line, and C the point given out of it, from which its required to let fall a perpendicular upon the indefinite given right line AB.

Affume any point D, on the oppofite fide of the right line AB; about the center C, with the diftance CD, defcribe a circle EDG; bifect EG in H, join CG, CH, CE; then CH is the perpendicular required.

B

For

« ΠροηγούμενηΣυνέχεια »