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PROP. XII. P R O B.
O draw a straight line perpendicular to a given ftraight 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 line perpendicular to AB from the point C.
Take any point D upon the other fide of AB, and from the centre C, at the diftance
a Io. Def.
CD, defcribe the circle EGF
fect FG in H, and join CF,
e Io. I.
CH, CG; the straight line CH, drawn from the given point C,
is perpendicular to the given ftraight line AB.
Because FH is equal to HG, and HC common to the two triangles FHC, GHC, the two fides FH, HC are equal to the
two GH, HC, each to each; and the base CF is equal to the de. Def. bafe CG; therefore the angle CHF is equal to the angle CHG; and they are adjacent angles; but when a ftraight line ftanding c. 8, 1. on a straight line makes the adjacent angles equal to one another, each of them is a right angle, and the ftraight line which ftands upon the other is called a perpendicular to it; therefore from the given point C a perpendicular CH has been drawn to the given ftraight line AB. Which was to be done.
PROP. XIII. THEOR.
HE angles which one straight line makes with another upon one fide 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 fide 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
b II. I.
C 2. Ax.
a Def. 1o. right angle; but, if not, from the point B draw BE at right angles to CD; therefore the angles CBE, EBD are two right angles; and because CBE is equal to the two angles CBA, ABE 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 thefe 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 demonftrated to be equal to the fame three angles; and things that are equal to the fame are equal 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 ftraight line, &c. Q. E. D.
F, at a point in a straight line, two other ftraight lines, upon the opposite fides of it, make the adjacent angles together equal to two right angles, thefe two ftraight lines fhall be in one and the fame ftraight line.
At the point B in the straight line AB, let the two ftraight lines BC, BD upon the oppofite fides of AB, make the adjacent angles ABC ABD equal together to two right angles. BD is in the fame ftraight line with CB.
For, if BD be not in the fame traight line with CB, let BE be C
in the fame ftraight line with it; therefore, because the ftraight Book I. line AB makes angles with the ftraight line CBE, upon one fide of it, the angles ABC, ABE are together equal to two a 13. 1. 1 right angles; but the angles ABC, ABD are likewife 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. Ax. angle ABD, the lefs to the greater, which is impoffible; therefore BE is not in the fame ftraight line with BC. And, in like manner, it may be demonftrated, that no other can be in the fame ftraight line with it but BD, which therefore is in the fame ftraight line with CB. Wherefore, if at a point, &c. Q. E. D.
two straight lines cut one another, the vertical, er oppofite, angles fhall be equal.
Let two ftraight lines AB, CD cut one another in the point E; the angle AEC fhall be equal to the angle DEB, and CEB to AED.
Because the ftraight 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 A
AED, DEB, thefe alfo 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 fame b 3. Ax, manner it can be demonftrated that the angles CEB, AED are equal. Therefore, if two ftraight lines, &c. Q. E. D.
COR. I. From this it is manifeft, that, if two ftraight lines cut one another, the angles they make at the point where they cut, are together equal to four right angles.
COR. 2. And confequently that all the angles made by any number of lines meeting in one point are together equal to four right angles.
a Io. I.
e 4. I.
d 15. I.
a 16. I.
one fide of a triangle be produced, the exterior angle is greater than either of the interior oppofite angles.
Let ABC be a triangle, and let its fide BC be produced to D, the exterior angle ACD is greater than either of the interior oppofite angles CBA, BAC.
Bifecta AC in E, join BE
Becaufe AE is equal to
remaining angles to the remaining angles, each to each, to which the equal fides are oppofite; wherefore the angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle EC; therefore the angle ACD is greater than BAE: In the fame manner, if the fide BC be bifected, it may be demonstrated that the angle BCG, that is, the angle ACD, is greater than the angle ABC. Therefore, if one fide, &c. Q. E. D.
PROP. XVII. THEOR.
NY two angles of a triangle are together less than two right angles.
Let ABC be any triangle;
Produce BC to D; and be-
thefe add the angle ACB; therefore the angles ACD, ACB are
PROP. XVIII. THEOR.
THE greater fide of every triangle is oppofite to the
Let ABC be a triangle, of which the fide AC is greater than the fide AB; the angle ABC is alfo greater than the angle BCA.
Because AC is greater than AB, make AD equal to AB, and join BD; and because ADB is the exterior angle of the triangle BDC, it is greater than
b 16. I.
the interior and oppofite angle DCB; but ADB is equal to c s. I
PRO P. XIX. THEOR.
THE greater angle of every triangle is fubtended by
Let ABC be a triangle, of which the angle ABC is greater
For, if it be not greater, AC
a s. 1