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3. Let it be granted that a circle may be described from any centre, and with

any

radius.

AXIOMS.

1. Things that are equal to the same thing are equal to each other.

2. If equals be added to equals, the sums are equals.

3. If equals be taken from cquals, the remainders are equals.

4. If equals be added to unequals, the sums are unequals.

5. If equals be taken from unequals, the remainders are unequals.

6 Things which are double of the same, or equal things, are equal to one another.

7. Things which are halves of the same, or equal things, are equal to one another.

8. Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another.

9. The whole is greater than its part, and equal to all its parts taken together.

10. Two straight lines cannot inclose a space. 11. All right angles are equal to one another.

12. If two magnitudes be equal, and one of them be greater than a third, the other is also greater than the third.

13. If two quantities be equal, and one of them be less than a third, the other is also less than the third.

14. If there are three magnitudes, such that the first is greater than the second, and the second greater than the third, much more is the first greater than the third.

15. If there are three magnitudes, such that the first is less than the second, and the second less than the third, much more is the first less than the third.

16. Through the same point there cannot be drawn two straight lines parallel to the same straight line without coinciding.

EXPLANATION OF SYMBOLS. L means angle.

means parallelogram. Ls angles.

straight line. 2L

right angle. at Ls right angles.

less than. equal to.

perpendicular. parallel to.

because. triangle.

therefore. triangles.

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greater than.

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رز

A8

PROPOSITION I.-THEOREM.

E

D

B

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The angles ACD and DCB, which one straight line, DC, makes with another, AB, on one side of it, are either two right angles, or are together equal to two right angles. A

If the LS ACD and DČB be equal, each of them is a r4L, (Def. 9.); but if they are not equal, conceive CE to be drawnt to AB, then the Ls ACE and ECB are two 7 Ls; but the three Is ACE, ECD, and DCB, are together = the two Ls ACE and ECB, and also to the two Ls ACD and DCB; .. the two Ls ACD and DCB are together = the two Ls ACE and ECB, but the two Ls ACE and ECB are two w Ls; .. the two Ls ACD and DCB are together equal to two r* Ls. Q. E. D.

Cor. 1. All the angles that can be formed at the point C, in the straight line

AB, on one side of it, are together equal to two right angles.

For the L XCD is = the two Ls ACE and ECD, so that the sum of the Ls is not increased by drawing the | EC, and in the same manner it may be shown the sum of the Ls would not be increased by drawing any number of Is through the point C.

Cor. 2. All the angles formed at the point C, on the other side of the line, by any number of lines meeting in C, will also be equal to two right angles.

Cor. 3. Hence all the angles formed round a point, by any number of lines meeting in it, are together equal to two right angles.

SCHOLIUM. For the purposes of calculation, the circumference of every circle is supposed to be divided into 360 equal parts, called degrees, and each degree is supposed to be divided into 60 equal parts, called minutes, and each minute into 60 equal parts, called seconds. Degrees, minutes, and seconds, are distinguished by the following marks: 703 24", which is read 7 degrees, 3 minutes, and 24 seconds.

In the same manner all the angles round about a point are' divided into the same number of degrees, minutes, and seconds. Since then the circle entirely surrounds its centre, and is similarly situated to it in every direction, the portion of the circumference intercepted between two lines drawn from the centre to the circumference, is the measure of the angle

B

с.

A

at the centre; thus the angle AOB is measured by the intercepted arc AB, and the angle COB is measured by the arc CB.

Since all the angles round a point are (Cor. 3.) equal together to four right angles, and also to 360°, the numerical measure of a right angle is 90°.

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PROPOSITION II.-THEOREM. If, at a point B, in a straight line AB, two

A
other straight lines, CB
and BD, on opposite sides
of AB, make the adjacent
angles ABC and ABD
together equal to two

С
B

D
right angles, these two
straight lines are in one and the same straight line.

For if BD be not in the same | with CB, let BE be in the same | with it; then since CBE is a l, and AB makes Ls with it, the two Ls ABC and ABE are together

- two of Ls; but the two Ls ABC and ABD are also together two r*Ls by supposition; the two Ls ABC and ABE are = the two Ls CBA and ABD; take away the common angle ABC, and there remains the LABE = the LABD, the less = the greater, which is impossible ; .. BE is not in the same / with CB, and in the same manner it can be shown that no | can be in the same with CB, except BD, which therefore is in the same with it. Q. E, D,

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D

E

B

PROPOSITION III.-THEOREM. If two straight lines, AB and CD, cut each other in the point E, the vertical or opposite angles, AEC, DEB, are equal.

C For the two Ls AEC and AED, which the | AE makes with the | CD, are together equal to two y Ls ; and the two Ls AED and DEB, which the | DE makes with the AB, are also together equal to two ad L8; .. the two Ls AEC and AED, are together equal to the two Ls AED and DEB; take from each the common LAED, and there remains the LAEC=DEB; in the same manner it may be demonstrated that the two L8 AED and CEB are equal.

Q: E. D.

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B

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D

PROPOSITION IV.-PROBLEM. To make a triangle AEB, whose three sides shall be

E equal to the three given straight lines, AB, C, and D.

F A

G From the centre A, with a radius equal to C, describe the circle EFH, (Post. 3), and from the centre B, with a radius equal to D, describe a circle EGH; and from the point E, where the circles cut each other, draw the \s AE and EB, (Post. 1), then AEB will be the triangle required.

Because AE is the radius of the circle EFH, and it was described with a radius =C, .. AE is =C, and because BE is the radius of the circle EGH, and it was described with a radius =D, ... EB is = D. Hence the A AEB has its three sides = to the three (s AB, C, and D.

Cor. 1. If the ls C and D were equal, the triangle would be isosceles.

Cor. 2. If the given lines, AB, C, and D, were all equal, the triangle would be equilateral.

Cor. 3. If Cʻand D were together less than AB, the circles would not intersect, and the construction would be impossible; hence any two sides of a triangle are together greater than the third, which is established in a different manner in (Prop. 13.)

PROPOSITION V.—THEOREM. If two triangles, ABC and DEF, have two sides, AC and CB, and the contained angle ACB in one, respectively equal to two sides, DF and FE, and the contained angle DFE in the other, the triangles are equal in every respect.

For conceive the point C to be laid on the point F, and the CA on the FD, then, .: these lines are =, A will coincide with the point D. And since CA coincides with FD, and the LC is = the LF, the line CB will fall on the line FE; and .:: CB and FE are =, the point B will coincide with the point E. Hence since

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the point

E

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the points A and B coincide with the points D and E, the line AB will coincide with the line DE, and the A ABC will coincide with the DEF; .. the two as are equal, and have all the parts of the one equal to the corresponding parts of the other, namely the side AB = the side DE, the LA = the LD, and the LB = the LE.

Q. E. D. PROPOSITION VI.-THEOREM. In any two triangles, ABC, DEF, if two angles, A and B in the one, be respectively equal to two angles, D and E in the other, and the sides AB and DE, which lie between these equal angles be also equal, the triangles are equal in all respects.

For conceive the point A to be laid on the point D, and the side AB on the side DE, then, ::: these lines are equal, the point B will

SBD coincide with the point E. And as AB and DE coincide, and the LA is = the LD, the side AC will coincide with DF, and for a like reason BC will coincide with EF. i. since AC falls on DF, and BC on EF, the point C must coincide with the point F; and :. the two os are in all respects equal, having the other sides, AC, BC, = the two, DF, EF, and the remaining LC = the remaining LF.

Q. E. D. PROPOSITION VII.-THEOREM. In an isosceles triangle, ABC, the angles A and B, opposite the equal sides BC and AC, are equal.

For, conceive the LC to be bisected by the line CD, then the two AS ACD and BCD have AC=BC, and CD common to both, and the LACD = the LBCD; .. they are equal in every respect, (Prop. 5), and have the LA = the LB, :;: they are opposite to the common side CD.

Q. E. D. Cor. 1. If the equal sides AC and BC be produced to E and F, the angles EAB and ABF on the other side of the base will also be equal. For, the two Ls CAB, BAE, are together

= two gt Ls, (Prop. 1), and the two Ls CBA and ABF are also together = to two he Ls; .. the two Ls CAB, BAE, are = the two Ls CBA, ABF, (Ax. 1); and it was proved that the

A

B В

D

E

F

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