Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Book I.

PROPOSITION I. PROBLEM.

[ocr errors]

describe an equilateral triangle upon a given right
line.

3.

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

About the center A, with the distance AB, describe the circle BCD *; and about the center B, with the distance BA, describe a Postulate the cirele ACEb; from the point C, where the two circles intersect each other, draw the lines CA, CBb.

b Post. 1. 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 BAC, but CA is proved equal to AB, and BC to AB; on 15. therefore BC is equal to AC d: Therefore the three sides AB, d Axiom 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.

PRO P. II. PROB.

A

T a given point to put a right line equal to a given

right line.

с

I.

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

With the center C and distance BC, describe the circle BGH';a Polt. 3. join ACb, upon which describe an equilateral triangle DACC; b Post

. s. produce DC that pafses through the center to G, in the circumference; and DA to any distance Ed; with the center D, and d Post. 2. distance DG, describe the circle KGL.

Then, because C is the center of the circle BGH, BCis equal to CG"; and because D is the center of the circle KGL, DG e Def. 15. is equal to DL o, but DC is equal to DAC: 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: 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.A

Book I.

PRO P. III. PRO B.

Wo unequal right lines being given, to cut of from the

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

From the point A draw a right line AD equal to C”, about b Post. 3. the center A, with the distance AD, describe a circle DEFb;

then, because A is the center of the circle DEF, AE is equal to c Def. 15. AD, but AD is equal to C: Therefore AE is likewise equal d Ax. I.

to Cd: Which was required.

PRO P. IV. THE ORE M.

[ocr errors]

F there be two triangles having two sides of the one equal to

two sides of the other, each to each ; and the angle contained by the two sides of the one, equal to the angle contained by the correspondent sides 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 sides subtend

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 bases 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 subtend.

For, let the triangle ABC be applied to the triangle DEF, so gas the point A may coincide with the point D, the right line JAB with DE, then the point B will coincide with the point E; for AB and DE are equal ; and, because the angles BAC 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 2 Def. 4.

lines BC and EF, will coincide"; and therefore equalo: There6 Ax. 8. fore the triangles ABC, DEF, are equal, the angle ABC e

qual to the angle DEF, and ACB to DFE, each to each, which the equal fides subtend. Wherefore, &c.

[ocr errors]

PRO P. V. THE O R.

Book I. HE angles above the base of every isosceles triangle are equal to one another.

to

to al

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

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

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

PRO P. VI. THEO R.

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

[merged small][ocr errors]

a

[ocr errors]

E;

th

gi ere

Let the triangle ABC have the angles ABC, ACB, equal; then the sides AB, AC, will likewise 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;
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, e-
qual ; therefore the base AB is equal to the base DC, and the
triangles ABC, DBC, equalb; that is, a part equal to the whole; b 4.
which is absurd : Wherefore the sides AB, AC, are equal ; that
is, the triangle is isosceles. Wherefore, &c.
Cor. Hence every equiangular triangle is also equilateral.

PRO P. VII. THE OR
[F from the extremity of any right line, to two different points

on the same side, there be drawn two right lines equal to one
another, the lines drawn from the other extremity, to the same
points, cannot be equal to one another.

If from the extremity A, of the right line AB, to the points
C,D, on the same 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.

Join DC; for, because AC, AD, are equal, the angles ADC,

ACD are equal a ; but the angle BDC is greater than ADCb, or b Ax. 9. ACD, and much greater than BCD, bat, if BD is equal to

BC, the angles BCD, BDC, are equal, and likewise greater ; which is impossible: Therefore, BD is not equal to BC. If BD is made equal to BC, it is proved in the same manner that AC is not equal to AD. Wherefore, &c.

a S.

[ocr errors]

PRO P. VIII. THEO R.
F two triangles have two sides of the one, equal to two sides of

the other, and the base of the one equal to the base of the other, then the angles.that these equal bases subtend will be equal to one another.

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

For, let the triangle ABC be applied to the triangle DEF, so 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 sides 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 impossible b. Wherefore the point A cannot fall in G, and, for the same reason, in no point but D; therefore the angle BAC is equal to the angle EDF. Wherefore, &c.

[merged small][ocr errors]

PRO P. IX. PROB.

TO

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

a 3.

bi.

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

Afsume any point D, in the right line AB, and cut off AE equal to AD *; join DE, upon which describe the equilateral triangle DEFb, and join AF; then the right line AF will bifect the given angle BAC; for, because DA is equal to EA, and AF is common, and the base DF equal to EFb; therefore the angle DAF is equal to the angle EAFC: Therefore the angle BAC is bisected by the right line AF. Which was requiredo

[ocr errors]

Book I.

PRO P. X. PRO B.

Tocut

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

Let AB be a given right line, required to be cut into two equal parts ; upon it describe an equilateral triangle ABC ; bifeat the angle ACB by the right line CD *; then is the right a g. line AB bifected in D.

For, because AC is equal to CB, and CD common, and the angles ACD, BCD, equal", the bases AD, BD, are equal b: b 4. Therefore the right line AB is b:fected in D. Which was required.

PRO P. XI. PRO B.

Teine given in the jare.

10 draw a line at right angles to a given right line from d

point given same.

[ocr errors]

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.

Affume any point D in AC, and make CE equal to CD2; 2 3. upon DE describe 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 sides DC, CF, are equal c. coník, to the two sides EC, CF, and the bases FD, FE, equal ), the angles FCD, FCE, are likewise equald: Therefore each of them d 8. is a right angle, and FC perpendicular to AB & Which was e Def. 104 required.

PRO P. XII. PRO B.

0 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.

Afsume any point D, on the opposite side of the right line
AB ; about the center C, with the distance CD, describe a circle
EDG; bisect EG in H, join CG, CH, CE; then CH is the
perpendicular required.
B

For

« ΠροηγούμενηΣυνέχεια »