Sidebilder
PDF
ePub
[ocr errors]

* 3 Post. * 10. 1.

+ Constr.

Let AB be the given straight line, which may be produced
to any length both ways, and let C be a point without it; it is
required to draw a straight line perpendi-
cular to AB, from the point C.

Take any point D upon the other side of
AB, and from the centre c, at the distance

AS
CD, describe * the circle EGF, meeting AB
in F, G; bisect * FG in H, and join Ch: the straight line
CH, drawn from the given point C, shall be perpendicular to
the given straight line AB.

Join CF, CG: and because FH is equal t to HG, and HC common to the two triangles FHC, GHC, the two sides FH, HC are equal to the two GH, HC, each to each ; and the base CF is equal* to the base CG: therefore the angle CHF is equal

the angle CHG ; and they are adjacent angles : but when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of them is a right angle, and the straight line, which stands upon the other, is called a perpendicular+ to it: therefore, from the given point c, a perpendicular ch has been drawn to the given straight line AB. Which was to be done.

15 Def. 8. 1.

+ 10 Def.

PROPOSITION XIII.

10 Def. * 11. 1. * 10 Def.

THEOR.The angles which one straight line makes with an

other upon one side of it, are either two right angles, or are together equal to two right angles.

Let the straight line AB make with CD, upon one side of it, the angles CBA, ABD: these shall either be two right angles, or shall together be equal to two right angles.

For if the angle CBA be equal to ABD, each of them is a right + angle: but if not, from the point B draw BE at right angles * to CD; therefore the angles CBE, EBD are two

А

E right angles and because the angle CBE is equal to the two angles CBA, ABE to- D с gether, add the angle EBD to each of these equals ; therefore the angles CBE, EBD are equal * to the three angles CBA, ABE, EBD. Again, because the angle DBA is equal to the two angles DBE, EBA, add to each of these equals the angle ABC; therefore the angles DBA, ABC are equalt to the three angles DBE, EBA, ABC: + 2 Ax. but the angles CBE, EBD have been demonstrated to be equal to the same three angles; and things that are equal to the same thing, are equal * to one another; therefore the angles * 1 Ax. CBE, EBD are equal to the angles DBA, ABC: but CBE, EBD are two right angles ; therefore, DBA, ABC are together equal + to two right angles. Wherefore, the angles which + 1 Ax. one straight line, &c. Q. E. D.

* 2 Ax.

PROPOSITION XIV.

A

E

THEOR.—If, at a point in a straight line, two other straight

lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line.

At the point B, in the straight line AB, let the two straight lines BC, BD upon the opposite side of AB, make the adjacent angles ABC, ABD equal together to two right angles: BD shall be in the same C

D straight line with CB.

For if BD be not in the same straight line with CB, let BE be in the same straight line with it: therefore, because the straight line AB makes with the straight line CBE, upon one side of it, the angles ABC, ABE, these angles are together equal * to two right angles; but the angles ABC, ABD • 13. 1, are likewise together equal t to two right angles; therefore, † Hyp. the angles CBA, ABE are equal t to the angles CBA, ABD: take + 1 Ax. away the common angle ABC, and the remaining angle ABE is equal * to the remaining angle ABD, the less to the greater, *3 Ax. which is impossible; therefore BE is not in the same straight line with BC. And in like manner, it may be demonstrated, that no other can be in the same straight line with it but BD, which, therefore, is in the same straight line with CB. Wherefore, if at a point, &c. Q. E. D.

PROPOSITION XV.

THEOR if iwo strairht lines cut one another, the vertical, or

opposite, angleó shali be equal. Let the 119 straight lines AB, CD cut one another in the point E: tre angle AEC shall be equal to the angle DEB, and CEB to AED.

* 13. 1.

[ocr errors]

* 13. 1.

Because the straight line AE makes with CD the angles CEA, AED, these angles are together equal to two right angles. Again, because the straight line DE makes with AB the angles AED, DEB, these also are together equal * to two right angles; and CEA, AED have been demonstrated to be A equal to two right angles ; wherefore the angles CEA, AED are equalt to the angles AED; DEB: take away the common angle AED, and the remaining angle CEA is equal * to the remaining angle DEB. In the same manner it can be demonstrated, that the angles CEB, AED are equal. Therefore, if two straight lines, &c.

+ 1 Ax.

* 3 Ax.

Q. E. D.

Cor. 1. From this it is manifest, that if two straight lines cut one another, the angles which they make at the point where they cut, are together equal to four right angles.

Cor. 2. And consequently, that all the angles made by any number of lines meeting in one point, are together equal to four right angles.

PROPOSITION XVI.

THEOR - If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.

10.1.

A

+ 3. 1.
† Constr.

• 15. 1.

B

Let ABC be a triangle, and let its side BC be produced to D: the exterior angle ACD shall be greater than either of the interior opposite angles CBA, BAC.

Bisect * AC in E ; join BE and produce it to F, and make
EF equal + to BE, and join FC.

F
Because AE is equal † to EC, and BE +
EF; AE, EB are equal to CE, EF, each

E
to each ; and the angle AEB is equal * to
the angle CEF, because they are opposite

D vertical angles ; therefore the base AB is

G equal" to the base CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to cach, to which the equal sides are opposite : wherefore the angle BAE is equal to the angle ECF: but the angle ECD is greater † than the angle ECF, therefore the angle ACD is greater than BAE.

In the same manner, if the side BC be bisected, and AC be produced to G, it may

[ocr errors][merged small]

be demonstrated, that the angle BCG, that is, the angle • • 15. 1. ACD, is greater than the angle ABC. Therefore, if one side, &c. Q. E. D.

• 16. 1.

PROPOSITION XVII.
THEOR.-Any two angles of a triangle are together less than

two right angles.
Let ABC be any triangle: any two of its angles together
shall be less than two right angles.

Produce BC to D: and because ACD is the exterior angle of the triangle ABC, ACD is greater * than the interior and opposite angle ABC; to each of these add the angle ACB ; therefore the angles ACD, ACB are greater +

B

+ 4 Ax. than the angles ABC, ACB : but ACD, ACB are together equal to two right angles; therefore the angles • 13. 1. ABC, BCA are less than two right angles. In like manner, it may be demonstrated, that BAC, ACB, as also CAB, ABC, are less than two right angles. Therefore, any two angles, &c. Q. E. D.

PROPOSITION XVIII.

THEOR.—The greater side of every triangle is opposite to the

greater angle.
Let ABC be a triangle, of which the side А
AC is greater than the side AB: the angle
ABC shall be greater than the angle BCA.
Because AC is greater than AB, make B

• 3. I. AD equal to AB, and join BD: and because ADB is the exterior angle of the triangle BDC, it is greater

• 16. 1. than the interior and opposite angle DCB ; but ADB is equal * • 5. 1. to ABD, because the side AB is equal + to the side AD; there- + Constr. fore the angle ABD is likewise greater than the angle ACB; therefore much more is the angle ABC greater than ACB. Therefore, the greater side, &c. Q. E. D.

PROPOSITION XIX.
THEOR.— The greater angle of every triangle is subtended by

the greater side, or has the greater side opposite to it.
Let ABC be a triangle, of which the angle ABC is greater

[ocr errors]

* 5.1.

+ Hyp.

18. 1.

B

Q. E. D.

See N.

than the angle BCA : the side AC shall be greater than the side AB.

For if it be not greater, AC must either be equal to AB, or
less than it: it is not equal, because then the angle ABC
would be equal * to the angle CAB ; but it
is t not; therefore AC is not equal to AB:
neither is it less, because then, the angle
ABC would be less * than the angle ACB;
but it is not; therefore the side AC is not
less than AB: and it has been shewn that
it is not equal to AB: therefore AC is greater than AB.
Wherefore, the greater angle, &c.

PROPOSITION XX.
THEOR.–Any two sides of a triangle are together greater

than the third side.
Let ABC be a triangle: any two sides of it together shall
be greater than the third side ; viz. the sides BA, AC greater
than the side BC; and AB, BC greater than AC; and BC, CA
greater than AB.
Produce BA to the point D, and make

D AD equal to AC; and join DC.

Because DA is equal to AC, the angle ADC is equal * to ACD; but the angle BCD is greater + than the angle ACD; therefore the angle BCD is greater than the angle ADC: and because the angle BCD of the triangle DCB, is greater than its angle BDC, and that the greater * angle is subtended by the greater side, therefore the side DB is greater than the side BC: but DB is equal to BA # and AC; therefore the sides BA, AC are greater than BC. In the same manner it may be demonstrated, that the sides AB, BC are greater than CA; and BC, CA greater than AB. Therefore, any two sides, &c. Q. E. D.

[ocr errors]

3.1.

* 5. 1.

B

+ 9 Ax.

* 19. 1.

PROPOSITION XXI.

See N.

Theor.-If from the ends of the side of a triangle, there be

drawn two straight lines to a point within the triangle, these
shall be less than the other two sides of the triangle, but shall
contain a greater angle.
Let ABC be a triangle, and from the points B, C, the ends

# Because AD is equal * to AC, add BA to each, therefore the whole BD is equalf to the two BA, AC.

* Constr.

+ 2 Ax

« ForrigeFortsett »