TO draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it.

C

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 perpendicular to AB from the point C.

sects FG in H, and join CF,

с 10. 1.

CH, CG; the straight line CH, drawn from the given point C,

is perpendicular to the given straight line AB.

1.

Because FH is equal 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 equald to the d 15. Def. base CG; therefore the angle CHF is equal to the angle CHG; and they are adjacent angles; but when a straight line standing e 8. 1. on a 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.

PROP. XIII. THEOR.

THE angles which one straight line makes with another upon the one side of it, are either two right angles, or are together equal to two right angles.

Book I.

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

For, if the angle CBA be equal to ABD each of them is a a Def.10. right a angle; but if not, from the point B draw BE at right

b 11. 1. angles b to CD; therefore the angles CBE, EBD are two right anglesa; and because CBE is equal to the two angles CBA, ABE c2. Ax. together, 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 these equals the angle ABC; therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC; but the angles CBE, EBD have been demonstrated to be equal to the same three angles; and things

d 1. Ax. that are equal to the same are equal d to one another; therefore the angles 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, when a straight line, &c. Q. E. D.

PROP. XIV. 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 sides of AB, make the adjacent angles ABC, ABD equal together to two right angles, BD is in the same straight line with CB.

For, if BD be not in the same straight line with CB, let BE be C

A

B

D

E

in the same straight line with it; therefore, because the straight Book I. line AB makes angles with the straight line CBE, upon one side of it, the angles ABC, ABE are together equal a to two right a 13. 14 angles; but the angles ABC, ABD are likewise together equal to two right angles; therefore the angles CBA, ABE are equal to the angles CBA, ABD: take away the common angle ABC, the remaining angle ABE is equal to the remaining angle b 3. Ax. ABD, the less to the greater, 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.

PROP. XV. THEOR.

IF two straight lines cut one another, the vertical, or opposite, angles shall be equal.

Let the two straight lines AB, CD cut one another in the point E; the angle AEC shall be equal to the angle DEB, and CEB to AED.

and CEA, AED have been de

monstrated to be equal to two right angles; wherefore the angles CEA, AED are equal to the angles AED, DEB. Take away the common angle AED, and the remaing angle CEA is equal b b 3. Ax. 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. Q. E. D.

COR. 1. From this it is manifest, that, if two straight lines cut one another, the angles 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.

D

Book I.

}

PROP. XVI. THEOR.

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

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

A

a 10. 1.

Bisecta AC in E, join BE and produce it to F, and make EF equal to BE; join also FC, and produce AC to

G.

E

F

AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, 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, it may be demonstrated that the angle BCG, d 15. 1. that is d, the angle ACD, is greater than the angle ABC. Therefore, if one side, &c. Q. E. D.

PROP. 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 are less than two right angles.

Produce BC to D; and because ACD is the exterior angle of the triangle ABC, ACD is a 16. 1. greater a than the interior and oppósite angle ABC; to each of

B

A

C

D

these add the angle ACB; therefore the angles ACD, ACB are Book I. greater than the angles ABC, ACB; but ACD, ACB are together equal to two right angles; therefore the angles ABCС, Ь 13. 1. 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.

PROP. XVIII. THEOR.

THE greater side of every triangle is opposite to the greater angle.

Let ABC be a triangle, of which the side AC is greater than the side AB; the angle ABC is also greater than the angle BCA.

A

Because AC is greater than AB, make a AD equal to AB, and join BD; and because ADB is the ex

D

a 3. 1.

terior angle of the triangle BDC, B

C

it is greater than the interior and opposite angle DCB; but b 16. 1. ADB is equal to ADB, because the side AB is equal to the side c 5. 1.

AD; therefore the angle ABD is likewise greater than the angle

ACB; wherefore much more is the angle ABC greater than

ACB. Therefore, the greater side, &c. Q. E. D.

PROP. 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 than the angle BCA; the side AC is likewise 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 a to the angle ACB; but it is not; therefore AC is not equal to AB; neither is it less; because then the angle ABC would be less than the angle ACB; B

A

1

a 5. 1.

Сь 18. 1.