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a Io. Def.
line, the angle CBE is equal
E to the angle EBA; in the same manner, because ABD is a straight line, the angle DBE is equal to the angle EBA ; wherefore the angle DBE is equal to
D the angle CBE, the less to the greater ; which is impossible ; therefore two straight lines can A
B not have a common segment.
PRO P. XII. PROB. 10 draw a straight line perpendicular to a given
straight line of an unlimited length, from a given point without it.
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
С line perpendicular to AB from the point C. Take any point Dupon the
E other side of AB, and from the centre C, at the distance CD, describe the circle EGF
b 3. Porto meeting AB in F, G; and bi F
G B fectFG in H, and join CF, CH, CG; the straight line CH, drawn from the given point C, is perpendicular to the given straight line AB.
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 equal to the d 55. Def. base CG; therefore the angle CHF is equal to the angle CHG; and they are adjacent angles; but when a straight line standing c.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,
PRO P. XIII, THEOR.
other upon one side of it, are either two right
e IO. I.
Let the straight line AB make with CD, upon one fide of
b II. I.
C 2. Ax.
d i. Ax.
D B a Def. 1o. right angle; but, if not, from the point B draw BE at right
angles 6 to CD; therefore the angles CBE, EBD are two right
PRO P. XIV. THEOR.
upon the opposite fides of it, make the adjacent angles
For, if BD be not in the same
in the same straight line with it; therefore, because the straight Book I. line AB makes angles with the straight line CBE, upon one w fide of it, the angles ABC, ABE are together equal to two a 13. 1. right 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 b 3. Ar. angle ABD, the less to the greater, which is impossible ; therefore BE is not in the fame 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.
posite, angles shall be equal.
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
A DĚ makes with AB the angles
B AED, DEB, these also are together equal to two right angles; and CEA, AED have been demonstrated 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 remaining angle CEA is equal to the remaining angle DEB. In the same b manner it can be demonstrated that the angles CEB, AED are equal. Therefore, if two straight lines, &c. Q. E. D.
COR. I. 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 points are together equal to four right angles.
PRO P. XVI.
T H E O R.
2 10. 1.
angle is greater than either of the interior opposite angles.
Let ABC be a triangle, and let its fide BC be produced to D,
Because AE is equal to
D cause they are opposite vertical angles; therefore the base AB is equal to the base CF, and the triangle AEB to
G the triangle CEF, and the remaining angles to the remaining angles, each to each, to which the equal fides are opposite; wherefore the angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle ECH; therefore the angle AČD is greater than BAE : In the same manner, if the side BC be bisected, it may be demonstrated that the angle BCG, that is 4, the angle ACD, is greater than the angle ABC. Therefore, if one fide, &c. Q. E. D.
NY two angles of a triangle are together less than
Produce BC to D; and be-
2 16. I.
these add the angle ACB ; therefore the angles ACD, ACB are Book ti greater than the angles ABC, ACB; but ACD, ACB are together equal to two right angles; therefore the angles ABC 6 13. I. BCA are less than two right angles. In like manner, it
may demonstrated that BAC, ACB, as also CAB, ABC are less than two right angles. Therefore any two angles, &c. Q. E. D.
PRO P. XVIII. THEOR.
B angle BDC, it is greater than
b 16. I. the interior and opposite angle DCB ; but ADB is equal to c 5.1 ABD, because the side AB is equal to the side 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 fide, &c. Q. E. D.
a 3. T.
PRO P. XIX. THE O R.
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 fide AC is likewise greater than the fide AB.
For, if it be not greater, AC muft either be equal to AB, or
А leis than it; it is not equal, because then the angle ABC would be equata to the angle ACB;
a si sa but it is not; therefore AC is not equal to AB; neither is it less; because then the angle B